Class XII Chapter 9 – Differential Equations Maths Question...
Transcript of Class XII Chapter 9 – Differential Equations Maths Question...
Class XII Chapter 9 – Differential Equations Maths
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Exercise 9.1
Question 1:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is four.
The given differential equation is not a polynomial equation in its derivatives. Hence, its
degree is not defined.
Question 2:
Determine order and degree(if defined) of differential equation
Answer
The given differential equation is:
The highest order derivative present in the differential equation is . Therefore, its order
is one.
It is a polynomial equation in . The highest power raised to is 1. Hence, its degree is
one.
Question 3:
Determine order and degree(if defined) of differential equation
Answer
Class XII Chapter 9 – Differential Equations Maths
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The highest order derivative present in the given differential equation is . Therefore,
its order is two.
It is a polynomial equation in and . The power raised to is 1.
Hence, its degree is one.
Question 4:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the given differential equation is . Therefore,
its order is 2.
The given differential equation is not a polynomial equation in its derivatives. Hence, its
degree is not defined.
Question 5:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is two.
Class XII Chapter 9 – Differential Equations Maths
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It is a polynomial equation in and the power raised to is 1.
Hence, its degree is one.
Question 6:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is three.
The given differential equation is a polynomial equation in .
The highest power raised to is 2. Hence, its degree is 2.
Question 7:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is three.
It is a polynomial equation in . The highest power raised to is 1. Hence, its
degree is 1.
Question 8:
Determine order and degree(if defined) of differential equation
Answer
Class XII Chapter 9 – Differential Equations Maths
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The highest order derivative present in the differential equation is . Therefore, its order
is one.
The given differential equation is a polynomial equation in and the highest power
raised to is one. Hence, its degree is one.
Question 9:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is two.
The given differential equation is a polynomial equation in and and the highest
power raised to is one.
Hence, its degree is one.
Question 10:
Determine order and degree(if defined) of differential equation
Answer
The highest order derivative present in the differential equation is . Therefore, its
order is two.
This is a polynomial equation in and and the highest power raised to is one.
Hence, its degree is one.
Question 11:
The degree of the differential equation
is
Class XII Chapter 9 – Differential Equations Maths
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(A) 3 (B) 2 (C) 1 (D) not defined
Answer
The given differential equation is not a polynomial equation in its derivatives. Therefore,
its degree is not defined.
Hence, the correct answer is D.
Question 12:
The order of the differential equation
is
(A) 2 (B) 1 (C) 0 (D) not defined
Answer
The highest order derivative present in the given differential equation is . Therefore,
its order is two.
Hence, the correct answer is A.
Class XII Chapter 9 – Differential Equations Maths
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Exercise 9.2
Question 1:
Answer
Differentiating both sides of this equation with respect to x, we get:
Now, differentiating equation (1) with respect to x, we get:
Substituting the values of in the given differential equation, we get the L.H.S.
as:
Thus, the given function is the solution of the corresponding differential equation.
Question 2:
Answer
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
L.H.S. = = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
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Question 3:
Answer
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
L.H.S. = = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Question 4:
Answer
Differentiating both sides of the equation with respect to x, we get:
Class XII Chapter 9 – Differential Equations Maths
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L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Question 5:
Answer
Differentiating both sides with respect to x, we get:
Substituting the value of in the given differential equation, we get:
Hence, the given function is the solution of the corresponding differential equation.
Question 6:
Answer
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Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
Hence, the given function is the solution of the corresponding differential equation.
Question 7:
Answer
Differentiating both sides of this equation with respect to x, we get:
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L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Question 8:
Answer
Differentiating both sides of the equation with respect to x, we get:
Substituting the value of in equation (1), we get:
Class XII Chapter 9 – Differential Equations Maths
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Hence, the given function is the solution of the corresponding differential equation.
Question 9:
Answer
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
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Hence, the given function is the solution of the corresponding differential equation.
Question 10:
Answer
Differentiating both sides of this equation with respect to x, we get:
Substituting the value of in the given differential equation, we get:
Hence, the given function is the solution of the corresponding differential equation.
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Question 11:
The numbers of arbitrary constants in the general solution of a differential equation of
fourth order are:
(A) 0 (B) 2 (C) 3 (D) 4
Answer
We know that the number of constants in the general solution of a differential equation
of order n is equal to its order.
Therefore, the number of constants in the general equation of fourth order differential
equation is four.
Hence, the correct answer is D.
Question 12:
The numbers of arbitrary constants in the particular solution of a differential equation of
third order are:
(A) 3 (B) 2 (C) 1 (D) 0
Answer
In a particular solution of a differential equation, there are no arbitrary constants.
Hence, the correct answer is D.
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Exercise 9.3
Question 1:
Answer
Differentiating both sides of the given equation with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Hence, the required differential equation of the given curve is
Question 2:
Answer
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Class XII Chapter 9 – Differential Equations Maths
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Dividing equation (2) by equation (1), we get:
This is the required differential equation of the given curve.
Question 3:
Answer
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Multiplying equation (1) with (2) and then adding it to equation (2), we get:
Now, multiplying equation (1) with equation (3) and subtracting equation (2) from it, we
get:
Substituting the values of in equation (3), we get:
Class XII Chapter 9 – Differential Equations Maths
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This is the required differential equation of the given curve.
Question 4:
Answer
Differentiating both sides with respect to x, we get:
Multiplying equation (1) with equation (2) and then subtracting it from equation (2), we
get:
Differentiating both sides with respect to x, we get:
Dividing equation (4) by equation (3), we get:
This is the required differential equation of the given curve.
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Question 5:
Answer
Differentiating both sides with respect to x, we get:
Again, differentiating with respect to x, we get:
Adding equations (1) and (3), we get:
This is the required differential equation of the given curve.
Question 6:
Form the differential equation of the family of circles touching the y-axis at the origin.
Answer
The centre of the circle touching the y-axis at origin lies on the x-axis.
Let (a, 0) be the centre of the circle.
Since it touches the y-axis at origin, its radius is a.
Now, the equation of the circle with centre (a, 0) and radius (a) is
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Differentiating equation (1) with respect to x, we get:
Now, on substituting the value of a in equation (1), we get:
This is the required differential equation.
Question 7:
Form the differential equation of the family of parabolas having vertex at origin and axis
along positive y-axis.
Answer
The equation of the parabola having the vertex at origin and the axis along the positive
y-axis is:
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Differentiating equation (1) with respect to x, we get:
Dividing equation (2) by equation (1), we get:
This is the required differential equation.
Question 8:
Form the differential equation of the family of ellipses having foci on y-axis and centre at
origin.
Answer
The equation of the family of ellipses having foci on the y-axis and the centre at origin is
as follows:
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Differentiating equation (1) with respect to x, we get:
Again, differentiating with respect to x, we get:
Substituting this value in equation (2), we get:
This is the required differential equation.
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Question 9:
Form the differential equation of the family of hyperbolas having foci on x-axis and
centre at origin.
Answer
The equation of the family of hyperbolas with the centre at origin and foci along the x-
axis is:
Differentiating both sides of equation (1) with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
Substituting the value of in equation (2), we get:
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This is the required differential equation.
Question 10:
Form the differential equation of the family of circles having centre on y-axis and radius
3 units.
Answer
Let the centre of the circle on y-axis be (0, b).
The differential equation of the family of circles with centre at (0, b) and radius 3 is as
follows:
Differentiating equation (1) with respect to x, we get:
Substituting the value of (y – b) in equation (1), we get:
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This is the required differential equation.
Question 11:
Which of the following differential equations has as the general solution?
A.
B.
C.
D.
Answer
The given equation is:
Differentiating with respect to x, we get:
Again, differentiating with respect to x, we get:
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This is the required differential equation of the given equation of curve.
Hence, the correct answer is B.
Question 12:
Which of the following differential equation has as one of its particular solution?
A.
B.
C.
D.
Answer
The given equation of curve is y = x.
Differentiating with respect to x, we get:
Again, differentiating with respect to x, we get:
Now, on substituting the values of y, from equation (1) and (2) in each of
the given alternatives, we find that only the differential equation given in alternative C is
correct.
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Hence, the correct answer is C.
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Exercise 9.4
Question 1:
Answer
The given differential equation is:
Now, integrating both sides of this equation, we get:
This is the required general solution of the given differential equation.
Question 2:
Answer
The given differential equation is:
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Now, integrating both sides of this equation, we get:
This is the required general solution of the given differential equation.
Question 3:
Answer
The given differential equation is:
Now, integrating both sides, we get:
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This is the required general solution of the given differential equation.
Question 4:
Answer
The given differential equation is:
Integrating both sides of this equation, we get:
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Substituting these values in equation (1), we get:
This is the required general solution of the given differential equation.
Question 5:
Answer
The given differential equation is:
Integrating both sides of this equation, we get:
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Let (ex + e–x) = t.
Differentiating both sides with respect to x, we get:
Substituting this value in equation (1), we get:
This is the required general solution of the given differential equation.
Question 6:
Answer
The given differential equation is:
Integrating both sides of this equation, we get:
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This is the required general solution of the given differential equation.
Question 7:
Answer
The given differential equation is:
Integrating both sides, we get:
Substituting this value in equation (1), we get:
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This is the required general solution of the given differential equation.
Question 8:
Answer
The given differential equation is:
Integrating both sides, we get:
This is the required general solution of the given differential equation.
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Question 9:
Answer
The given differential equation is:
Integrating both sides, we get:
Substituting this value in equation (1), we get:
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This is the required general solution of the given differential equation.
Question 10:
Answer
The given differential equation is:
Integrating both sides, we get:
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Substituting the values of in equation (1), we get:
This is the required general solution of the given differential equation.
Question 11:
Answer
The given differential equation is:
Integrating both sides, we get:
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Comparing the coefficients of x2 and x, we get:
A + B = 2
B + C = 1
A + C = 0
Solving these equations, we get:
Substituting the values of A, B, and C in equation (2), we get:
Therefore, equation (1) becomes:
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Substituting C = 1 in equation (3), we get:
Question 12:
Answer
Integrating both sides, we get:
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Comparing the coefficients of x2, x, and constant, we get:
Solving these equations, we get
Substituting the values of A, B, and C in equation (2), we get:
Therefore, equation (1) becomes:
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Substituting the value of k2 in equation (3), we get:
Question 13:
Answer
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Integrating both sides, we get:
Substituting C = 1 in equation (1), we get:
Question 14:
Answer
Integrating both sides, we get:
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Substituting C = 1 in equation (1), we get:
y = sec x
Question 15:
Find the equation of a curve passing through the point (0, 0) and whose differential
equation is .
Answer
The differential equation of the curve is:
Integrating both sides, we get:
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Substituting this value in equation (1), we get:
Now, the curve passes through point (0, 0).
Substituting in equation (2), we get:
Hence, the required equation of the curve is
Question 16:
For the differential equation find the solution curve passing
through the point (1, –1).
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Answer
The differential equation of the given curve is:
Integrating both sides, we get:
Now, the curve passes through point (1, –1).
Substituting C = –2 in equation (1), we get:
This is the required solution of the given curve.
Question 17:
Find the equation of a curve passing through the point (0, –2) given that at any point
on the curve, the product of the slope of its tangent and y-coordinate of the point
is equal to the x-coordinate of the point.
Answer
Let x and y be the x-coordinate and y-coordinate of the curve respectively.
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We know that the slope of a tangent to the curve in the coordinate axis is given by the
relation,
According to the given information, we get:
Integrating both sides, we get:
Now, the curve passes through point (0, –2).
∴ (–2)2 – 02 = 2C
⇒ 2C = 4
Substituting 2C = 4 in equation (1), we get:
y2 – x2 = 4
This is the required equation of the curve.
Question 18:
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line
segment joining the point of contact to the point (–4, –3). Find the equation of the curve
given that it passes through (–2, 1).
Answer
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It is given that (x, y) is the point of contact of the curve and its tangent.
The slope (m1) of the line segment joining (x, y) and (–4, –3) is
We know that the slope of the tangent to the curve is given by the relation,
According to the given information:
Integrating both sides, we get:
This is the general equation of the curve.
It is given that it passes through point (–2, 1).
Substituting C = 1 in equation (1), we get:
y + 3 = (x + 4)2
This is the required equation of the curve.
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Question 19:
The volume of spherical balloon being inflated changes at a constant rate. If initially its
radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t
seconds.
Answer
Let the rate of change of the volume of the balloon be k (where k is a constant).
Integrating both sides, we get:
⇒ 4π × 33 = 3 (k × 0 + C)
⇒ 108π = 3C
⇒ C = 36π
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At t = 3, r = 6:
⇒ 4π × 63 = 3 (k × 3 + C)
⇒ 864π = 3 (3k + 36π)
⇒ 3k = –288π – 36π = 252π
⇒ k = 84π
Substituting the values of k and C in equation (1), we get:
Thus, the radius of the balloon after t seconds is .
Question 20:
In a bank, principal increases continuously at the rate of r% per year. Find the value of r
if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).
Class XII Chapter 9 – Differential Equations Maths
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Answer
Let p, t, and r represent the principal, time, and rate of interest respectively.
It is given that the principal increases continuously at the rate of r% per year.
Integrating both sides, we get:
It is given that when t = 0, p = 100.
⇒ 100 = ek … (2)
Now, if t = 10, then p = 2 × 100 = 200.
Therefore, equation (1) becomes:
Class XII Chapter 9 – Differential Equations Maths
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Hence, the value of r is 6.93%.
Question 21:
In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs
1000 is deposited with this bank, how much will it worth after 10 years .
Answer
Let p and t be the principal and time respectively.
It is given that the principal increases continuously at the rate of 5% per year.
Integrating both sides, we get:
Now, when t = 0, p = 1000.
⇒ 1000 = eC … (2)
At t = 10, equation (1) becomes:
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Hence, after 10 years the amount will worth Rs 1648.
Question 22:
In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours.
In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is
proportional to the number present?
Answer
Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number present.
Integrating both sides, we get:
Let y0 be the number of bacteria at t = 0.
⇒ log y0 = C
Substituting the value of C in equation (1), we get:
Also, it is given that the number of bacteria increases by 10% in 2 hours.
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Substituting this value in equation (2), we get:
Therefore, equation (2) becomes:
Now, let the time when the number of bacteria increases from 100000 to 200000 be t1.
⇒ y = 2y0 at t = t1
From equation (4), we get:
Hence, in hours the number of bacteria increases from 100000 to 200000.
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Question 23:
The general solution of the differential equation
A.
B.
C.
D.
Answer
Integrating both sides, we get:
Hence, the correct answer is A.
Class XII Chapter 9 – Differential Equations Maths
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Exercise 9.5
Question 1:
Answer
The given differential equation i.e., (x2 + xy) dy = (x2 + y2) dx can be written as:
This shows that equation (1) is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Differentiating both sides with respect to x, we get:
Substituting the values of v and in equation (1), we get:
Class XII Chapter 9 – Differential Equations Maths
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Integrating both sides, we get:
This is the required solution of the given differential equation.
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Question 2:
Answer
The given differential equation is:
Thus, the given equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Differentiating both sides with respect to x, we get:
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:
Class XII Chapter 9 – Differential Equations Maths
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This is the required solution of the given differential equation.
Question 3:
Answer
The given differential equation is:
Thus, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Class XII Chapter 9 – Differential Equations Maths
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Integrating both sides, we get:
This is the required solution of the given differential equation.
Question 4:
Answer
The given differential equation is:
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Class XII Chapter 9 – Differential Equations Maths
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Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:
This is the required solution of the given differential equation.
Question 5:
Answer
The given differential equation is:
Class XII Chapter 9 – Differential Equations Maths
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Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:
Class XII Chapter 9 – Differential Equations Maths
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This is the required solution for the given differential equation.
Question 6:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Class XII Chapter 9 – Differential Equations Maths
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Substituting the values of v and in equation (1), we get:
Integrating both sides, we get:
This is the required solution of the given differential equation.
Question 7:
Answer
The given differential equation is:
Class XII Chapter 9 – Differential Equations Maths
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Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
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Integrating both sides, we get:
This is the required solution of the given differential equation.
Class XII Chapter 9 – Differential Equations Maths
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Question 8:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Class XII Chapter 9 – Differential Equations Maths
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Integrating both sides, we get:
This is the required solution of the given differential equation.
Question 9:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Class XII Chapter 9 – Differential Equations Maths
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Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:
Class XII Chapter 9 – Differential Equations Maths
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Therefore, equation (1) becomes:
This is the required solution of the given differential equation.
Question 10:
Answer
Class XII Chapter 9 – Differential Equations Maths
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Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
x = vy
Substituting the values of x and in equation (1), we get:
Integrating both sides, we get:
Class XII Chapter 9 – Differential Equations Maths
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This is the required solution of the given differential equation.
Question 11:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Class XII Chapter 9 – Differential Equations Maths
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Integrating both sides, we get:
Now, y = 1 at x = 1.
Substituting the value of 2k in equation (2), we get:
This is the required solution of the given differential equation.
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Question 12:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Class XII Chapter 9 – Differential Equations Maths
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Integrating both sides, we get:
Now, y = 1 at x = 1.
Substituting in equation (2), we get:
This is the required solution of the given differential equation.
Question 13:
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Answer
Therefore, the given differential equation is a homogeneous equation.
To solve this differential equation, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:
Class XII Chapter 9 – Differential Equations Maths
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Now, .
Substituting C = e in equation (2), we get:
This is the required solution of the given differential equation.
Question 14:
Answer
Therefore, the given differential equation is a homogeneous equation.
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To solve it, we make the substitution as:
y = vx
Substituting the values of y and in equation (1), we get:
Integrating both sides, we get:
This is the required solution of the given differential equation.
Now, y = 0 at x = 1.
Substituting C = e in equation (2), we get:
This is the required solution of the given differential equation.
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Question 15:
Answer
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
Substituting the value of y and in equation (1), we get:
Integrating both sides, we get:
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Now, y = 2 at x = 1.
Substituting C = –1 in equation (2), we get:
This is the required solution of the given differential equation.
Question 16:
A homogeneous differential equation of the form can be solved by making the
substitution
A. y = vx
B. v = yx
C. x = vy
D. x = v
Answer
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For solving the homogeneous equation of the form , we need to make the
substitution as x = vy.
Hence, the correct answer is C.
Question 17:
Which of the following is a homogeneous differential equation?
A.
B.
C.
D.
Answer
Function F(x, y) is said to be the homogenous function of degree n, if
F(λx, λy) = λn F(x, y) for any non-zero constant (λ).
Consider the equation given in alternativeD:
Hence, the differential equation given in alternative D is a homogenous equation.
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Exercise 9.6
Question 1:
Answer
The given differential equation is
This is in the form of
The solution of the given differential equation is given by the relation,
Class XII Chapter 9 – Differential Equations Maths
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Therefore, equation (1) becomes:
This is the required general solution of the given differential equation.
Question 2:
Answer
The given differential equation is
The solution of the given differential equation is given by the relation,
This is the required general solution of the given differential equation.
Question 3:
Answer
The given differential equation is:
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The solution of the given differential equation is given by the relation,
This is the required general solution of the given differential equation.
Question 4:
Answer
The given differential equation is:
The general solution of the given differential equation is given by the relation,
Question 5:
Answer
Class XII Chapter 9 – Differential Equations Maths
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By second fundamental theorem of calculus, we obtain
Question 6:
Answer
The given differential equation is:
This equation is in the form of a linear differential equation as:
The general solution of the given differential equation is given by the relation,
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Question 7:
Answer
The given differential equation is:
This equation is the form of a linear differential equation as:
The general solution of the given differential equation is given by the relation,
Class XII Chapter 9 – Differential Equations Maths
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Substituting the value of in equation (1), we get:
This is the required general solution of the given differential equation.
Question 8:
Answer
This equation is a linear differential equation of the form:
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The general solution of the given differential equation is given by the relation,
Question 9:
Answer
This equation is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
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Question 10:
Answer
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
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Question 11:
Answer
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
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Question 12:
Answer
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
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Question 13:
Answer
The given differential equation is
This is a linear equation of the form:
The general solution of the given differential equation is given by the relation,
Now,
Therefore,
Substituting C = –2 in equation (1), we get:
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Hence, the required solution of the given differential equation is
Question 14:
Answer
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
Now, y = 0 at x = 1.
Therefore,
Class XII Chapter 9 – Differential Equations Maths
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Substituting in equation (1), we get:
This is the required general solution of the given differential equation.
Question 15:
Answer
The given differential equation is
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
Now,
Therefore, we get:
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Substituting C = 4 in equation (1), we get:
This is the required particular solution of the given differential equation.
Question 16:
Find the equation of a curve passing through the origin given that the slope of the
tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the
point.
Answer
Let F (x, y) be the curve passing through the origin.
At point (x, y), the slope of the curve will be
According to the given information:
This is a linear differential equation of the form:
The general solution of the given differential equation is given by the relation,
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Substituting in equation (1), we get:
The curve passes through the origin.
Therefore, equation (2) becomes:
1 = C
⇒ C = 1
Substituting C = 1 in equation (2), we get:
Hence, the required equation of curve passing through the origin is
Question 17:
Find the equation of a curve passing through the point (0, 2) given that the sum of the
coordinates of any point on the curve exceeds the magnitude of the slope of the tangent
to the curve at that point by 5.
Answer
Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent
to the curve at (x, y) is
According to the given information:
This is a linear differential equation of the form:
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The general equation of the curve is given by the relation,
Therefore, equation (1) becomes:
The curve passes through point (0, 2).
Therefore, equation (2) becomes:
0 + 2 – 4 = Ce0
⇒ – 2 = C
⇒ C = – 2
Substituting C = –2 in equation (2), we get:
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This is the required equation of the curve.
Question 18:
The integrating factor of the differential equation is
A. e–x
B. e–y
C.
D. x
Answer
The given differential equation is:
This is a linear differential equation of the form:
The integrating factor (I.F) is given by the relation,
Hence, the correct answer is C.
Question 19:
The integrating factor of the differential equation.
is
A.
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B.
C.
D.
Answer
The given differential equation is:
This is a linear differential equation of the form:
The integrating factor (I.F) is given by the relation,
Hence, the correct answer is D.
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Miscellaneous Solutions
Question 1:
For each of the differential equations given below, indicate its order and degree (if
defined).
(i)
(ii)
(iii)
Answer
(i) The differential equation is given as:
The highest order derivative present in the differential equation is . Thus, its order is
two. The highest power raised to is one. Hence, its degree is one.
(ii) The differential equation is given as:
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The highest order derivative present in the differential equation is . Thus, its order is
one. The highest power raised to is three. Hence, its degree is three.
(iii) The differential equation is given as:
The highest order derivative present in the differential equation is . Thus, its order is
four.
However, the given differential equation is not a polynomial equation. Hence, its degree
is not defined.
Question 2:
For each of the exercises given below, verify that the given function (implicit or explicit)
is a solution of the corresponding differential equation.
(i)
(ii)
(iii)
(iv)
Answer
(i)
Differentiating both sides with respect to x, we get:
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Again, differentiating both sides with respect to x, we get:
Now, on substituting the values of and in the differential equation, we get:
⇒ L.H.S. ≠ R.H.S.
Hence, the given function is not a solution of the corresponding differential equation.
(ii)
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
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Now, on substituting the values of and in the L.H.S. of the given differential
equation, we get:
Hence, the given function is a solution of the corresponding differential equation.
(iii)
Differentiating both sides with respect to x, we get:
Again, differentiating both sides with respect to x, we get:
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Substituting the value of in the L.H.S. of the given differential equation, we get:
Hence, the given function is a solution of the corresponding differential equation.
(iv)
Differentiating both sides with respect to x, we get:
Substituting the value of in the L.H.S. of the given differential equation, we get:
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Hence, the given function is a solution of the corresponding differential equation.
Question 3:
Form the differential equation representing the family of curves given by
where a is an arbitrary constant.
Answer
Differentiating with respect to x, we get:
From equation (1), we get:
On substituting this value in equation (3), we get:
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Hence, the differential equation of the family of curves is given as
Question 4:
Prove that is the general solution of differential
equation , where c is a parameter.
Answer
This is a homogeneous equation. To simplify it, we need to make the substitution as:
Substituting the values of y and in equation (1), we get:
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Integrating both sides, we get:
Substituting the values of I1 and I2 in equation (3), we get:
Therefore, equation (2) becomes:
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Hence, the given result is proved.
Question 5:
Form the differential equation of the family of circles in the first quadrant which touch
the coordinate axes.
Answer
The equation of a circle in the first quadrant with centre (a, a) and radius (a) which
touches the coordinate axes is:
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Differentiating equation (1) with respect to x, we get:
Substituting the value of a in equation (1), we get:
Hence, the required differential equation of the family of circles is
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Question 6:
Find the general solution of the differential equation
Answer
Integrating both sides, we get:
Question 7:
Show that the general solution of the differential equation is given by
(x + y + 1) = A (1 – x – y – 2xy), where A is parameter
Answer
Integrating both sides, we get:
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Hence, the given result is proved.
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Question 8:
Find the equation of the curve passing through the point whose differential
equation is,
Answer
The differential equation of the given curve is:
Integrating both sides, we get:
The curve passes through point
On substituting in equation (1), we get:
Hence, the required equation of the curve is
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Question 9:
Find the particular solution of the differential equation
, given that y = 1 when x = 0
Answer
Integrating both sides, we get:
Substituting these values in equation (1), we get:
Now, y = 1 at x = 0.
Therefore, equation (2) becomes:
Substituting in equation (2), we get:
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This is the required particular solution of the given differential equation.
Question 10:
Solve the differential equation
Answer
Differentiating it with respect to y, we get:
From equation (1) and equation (2), we get:
Integrating both sides, we get:
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Question 11:
Find a particular solution of the differential equation , given that
y = – 1, when x = 0 (Hint: put x – y = t)
Answer
Substituting the values of x – y and in equation (1), we get:
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Integrating both sides, we get:
Now, y = –1 at x = 0.
Therefore, equation (3) becomes:
log 1 = 0 – 1 + C
⇒ C = 1
Substituting C = 1 in equation (3) we get:
This is the required particular solution of the given differential equation.
Question 12:
Solve the differential equation
Answer
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This equation is a linear differential equation of the form
The general solution of the given differential equation is given by,
Question 13:
Find a particular solution of the differential equation ,
given that y = 0 when
Answer
The given differential equation is:
This equation is a linear differential equation of the form
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The general solution of the given differential equation is given by,
Now,
Therefore, equation (1) becomes:
Substituting in equation (1), we get:
This is the required particular solution of the given differential equation.
Question 14:
Find a particular solution of the differential equation , given that y = 0
when x = 0
Answer
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Integrating both sides, we get:
Substituting this value in equation (1), we get:
Now, at x = 0 and y = 0, equation (2) becomes:
Substituting C = 1 in equation (2), we get:
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This is the required particular solution of the given differential equation.
Question 15:
The population of a village increases continuously at the rate proportional to the number
of its inhabitants present at any time. If the population of the village was 20000 in 1999
and 25000 in the year 2004, what will be the population of the village in 2009?
Answer
Let the population at any instant (t) be y.
It is given that the rate of increase of population is proportional to the number of
inhabitants at any instant.
Integrating both sides, we get:
log y = kt + C … (1)
In the year 1999, t = 0 and y = 20000.
Therefore, we get:
log 20000 = C … (2)
In the year 2004, t = 5 and y = 25000.
Therefore, we get:
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In the year 2009, t = 10 years.
Now, on substituting the values of t, k, and C in equation (1), we get:
Hence, the population of the village in 2009 will be 31250.
Question 16:
The general solution of the differential equation is
A. xy = C
B. x = Cy2
C. y = Cx
D. y = Cx2
Answer
The given differential equation is:
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Integrating both sides, we get:
Hence, the correct answer is C.
Question 17:
The general solution of a differential equation of the type is
A.
B.
C.
D.
Answer
The integrating factor of the given differential equation
The general solution of the differential equation is given by,
Hence, the correct answer is C.
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Question 18:
The general solution of the differential equation is
A. xey + x2 = C
B. xey + y2 = C
C. yex + x2 = C
D. yey + x2 = C
Answer
The given differential equation is:
This is a linear differential equation of the form
The general solution of the given differential equation is given by,
Hence, the correct answer is C.
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DIFFERENTIAL EQUATIONS
INTRODUCTION
1. Problems based on the order and degree of the differential equations
Working rule
(a) In order to find the order of a differential equations, see the highest derivative in the
given differential equation. Write down the order of this highest order derivatives.
(b) In order to find the degree of a differential equation write down the power of the
highest order derivative after making the derivatives occurring in the given differential
equation free from radicals and fractions.
2. Problems based on formation of differential equation.
Working rule
(a) Write the given equation.
(b) Differentiate the given equation w.r.t. independent variable of x as many times as the
number of arbitrary constants.
(c) Eliminate the arbitrary constants from given equation and the equations obtained by
differentiation.
3. Problems based on solution of differential equation in which variables are separable.
Working rule
This differential equation can be solved by the variable separable method which can be
put in the form
𝑑𝑦
𝑑𝑥 = 𝑓(𝑥). 𝑔(𝑦).
i.e., in which 𝑑𝑦
𝑑𝑥 can be expressed as the product of two functions, one of which is
a function of x only and the other a function of y only.
In order to solve the equation 𝑑𝑦
𝑑𝑥= 𝑓(𝑥). 𝑔(𝑦). Write down this equation in the
form 𝑑𝑦
𝑔(𝑦)= 𝑓(𝑥). 𝑑𝑥, then the solution will be ∫
𝑑𝑦
𝑔(𝑦)= ∫ 𝑓(𝑥)𝑑𝑥 + 𝑐, where C is an
arbitrary constant.
4. Problems based on solution of differential equations which are homogeneous.
Working rule
(a) Write down the given differential equation in the form 𝑑𝑦
𝑑𝑥 = 𝑓(𝑥, 𝑦)
(b) If 𝑓(𝑘𝑥, 𝑘𝑦) = 𝑓(𝑥, 𝑦). then differential equation is homogeneous.
(c) In order to solve, put 𝑦 = 𝑣𝑥, so that 𝑑𝑦
𝑑𝑥= 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥 and then separate the
variables𝑥 and 𝑣.
(d) Now solve the obtained differential equation by the variable separable
method. At the end put 𝑦
𝑥 in place of 𝑣.
5. Working Rule for Linear Differential Equation of first degree:
(a) Type 1.𝑑𝑦
𝑑𝑥+ 𝑃𝑦 = 𝑄 , where P and Q are constants or function of x only
General solution is𝑦. 𝐼𝐹 = ∫(𝑄. 𝐼𝐹)𝑑𝑥 + 𝐶 , where 𝐼𝐹 = 𝑒∫𝑃𝑑𝑥.
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(b) Type 2.𝑑𝑥
𝑑𝑦+ 𝑃𝑥 = 𝑄 , where P and Q are constants or function of y only
General solution is𝑥. 𝐼𝐹 = ∫(𝑄. 𝐼𝐹)𝑑𝑦 + 𝐶 , where 𝐼𝐹 = 𝑒∫𝑃𝑑𝑦.
Note: Particular solution can be obtained after getting the value of parameter C by
substituting the given initial values of the variables.
IMPORTANT SOLVED PROBLEMS
1. Solve the differential equation (x + y ) dy + ( x – y ) dx = 0
Solution
(x + y ) dy + ( x – y ) dx = 0
𝑑𝑦
𝑑𝑥 =
𝑦−𝑥
𝑥+𝑦
Put y = vx
𝑑𝑦
𝑑𝑥 = 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
∴ 𝑣 + 𝑥 𝑑𝑣
𝑑𝑥 =
𝑣𝑥−𝑥
𝑥 + 𝑣𝑥
𝑣 + 𝑥 𝑑𝑣
𝑑𝑥 =
𝑣−1
𝑣+1
𝑥 𝑑𝑣
𝑑𝑥 =
𝑣−1
𝑣+1 − 𝑣
𝑥 𝑑𝑣
𝑑𝑥 =
−( 1+ 𝑣2)
1+𝑣
1+𝑣
1+ 𝑣2 𝑑𝑣 = −𝑑𝑥
𝑥
Integrating both sides we get
∫1+𝑣
1+ 𝑣2 𝑑𝑣 = - ∫𝑑𝑥
𝑥
∫𝑑𝑣
1+ 𝑣2 + 1
2∫
2𝑣
1+𝑣2 𝑑𝑣 = − log 𝑥 + 𝑐
tan−1 𝑣 +1
2log|1 + 𝑣2| = − log 𝑥 + 𝑐
tan−1 𝑦
𝑥 +
1
2 𝑙𝑜𝑔 (1 +
𝑦2
𝑥2) = −𝑙𝑜𝑔 𝑥 + 𝑐
tan−1 𝑦
𝑥 +
1
2 𝑙𝑜𝑔 (𝑥2 + 𝑦2) = 𝑐
2. Solve 𝑥√1 − 𝑦2 𝑑𝑥 + 𝑦 √1 − 𝑥2 𝑑𝑦 = 0
Solution
𝑥√1 − 𝑦2 𝑑𝑥 = - 𝑦 √1 − 𝑥2 𝑑𝑦
𝑥
√1− 𝑥2𝑑𝑥 = −
𝑦
√1− 𝑦2𝑑𝑦
Integrating both sides
∫𝑥
√1− 𝑥2𝑑𝑥 = - ∫
𝑦
√1− 𝑦2𝑑𝑦
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−1
2∫
𝑑𝑡
√𝑡 =
1
2∫
𝑑𝑢
√𝑢 ( put t = 1 – x2 and put u = 1 – y2 )
− √𝑡 = √𝑢 + C
√1 − 𝑥2 + √ 1 − 𝑦2 = 𝐶
3. Solve the differential equation 𝑑𝑦
𝑑𝑥−
1
𝑥 . 𝑦 = 2𝑥2
Solution
The diff.eqn is in the form 𝑑𝑦
𝑑𝑥 + 𝑃𝑦 = 𝑄
Where P = −1
𝑥 and Q = 2x2
I.F = 𝑒∫𝑃 𝑑𝑥 = 𝑒∫−1
𝑥𝑑𝑥 = 𝑒−𝑙𝑜𝑔 𝑥 =
1
𝑥
Multiplying both sides of diff.eqn by I.F we get
1
𝑥
𝑑𝑦
𝑑𝑥 −
1
𝑥2𝑦 = 2𝑥
𝑑
𝑑𝑥(𝑦.
1
𝑥) = 2𝑥
Integrating both sides w.r.t.x we get
𝑦.1
𝑥 = 𝑥2 + 𝑐
y = x3 + Cx
4. Solve the diff. eqn 𝑥 𝑑𝑦
𝑑𝑥 = 𝑦 − 𝑥 tan
𝑦
𝑥
Solution
𝑑𝑦
𝑑𝑥 =
𝑦 − 𝑥 tan𝑦
𝑥
𝑥
Put y = vx =>𝑑𝑦
𝑑𝑥 = 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
∴ 𝑣 + 𝑥 𝑑𝑣
𝑑𝑥 =
𝑣𝑥 − 𝑥 𝑡𝑎𝑛𝑣
𝑥
𝑣 + 𝑥 𝑑𝑣
𝑑𝑥 = v – tan v
𝑥 𝑑𝑣
𝑑𝑥 = −𝑡𝑎𝑛 𝑣
cot v dv = - 𝑑𝑥
𝑥
Integrating both sides ,we get
∫ 𝑐𝑜𝑡 𝑣 𝑑𝑣 = − ∫𝑑𝑥
𝑥
log sin v = - log x + log C
log [x. sin (𝑦
𝑥) ] = 𝑙𝑜𝑔 𝑐
x sin (𝑦
𝑥) = C
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5. 𝑥𝑦 𝑑𝑦
𝑑𝑥 = (𝑥 + 2 )(𝑦 + 2 ) , find the equation of the curve passing through the points (1, -
1)
Solution
𝑥𝑦 𝑑𝑦
𝑑𝑥 = (𝑥 + 2 )(𝑦 + 2 )
𝑦
𝑦+2 𝑑𝑦 =
𝑥+2
𝑥 𝑑𝑥
(1 − 2
𝑦+2)𝑑𝑦 = (1 +
2
𝑥) 𝑑𝑥
Integrating both sides weget
Y –2 log (y + 2 ) = x +2 log x + C
The curve is pasing through the the point (1,-1)
- 1 –2 log 1 = 1 + 2 log 1 + C => C = -2
The equation of the line is 𝑦 − 𝑥 = 2 𝑙𝑜𝑔[x(y+2)] – 2
6. Solve the differential equation (𝑥2 − 1)𝑑𝑦
𝑑𝑥 + 2𝑥𝑦 =
2
𝑥2−1
Solution
(𝑥2 − 1)𝑑𝑦
𝑑𝑥 + 2𝑥𝑦 =
2
𝑥2 − 1
𝑑𝑦
𝑑𝑥+ (
2𝑥
𝑥2−1) . 𝑦 =
2
(𝑥2−1)2
I.F. = 𝑒∫2𝑥
𝑥2−1𝑑𝑥
= 𝑒𝑙𝑜𝑔(𝑥2−1) = (𝑥2 − 1)
𝑑
𝑑𝑥(𝑦 (𝑥2 − 1)) =
2
𝑥2−1
Integrating both sides w.r.t x we get
𝑦 (𝑥2 − 1) = ∫2
𝑥2−1 𝑑𝑥
𝑦 (𝑥2 − 1) = 𝑙𝑜𝑔 (𝑥−1
𝑥+1) + 𝐶
-
PRACTICE PROBLEMS
LEVEL I
1. Find the order and degree of the following differential equation.
(𝑑3𝑦
𝑑𝑥3)2
− 𝑥 (𝑑𝑦
𝑑𝑥)3
= 0
2. Show that 𝑦 = 𝐴𝑥 + 𝐵
𝑥 is a solution of the differential equation.
𝑥2 𝑑2𝑦
𝑑𝑥2 + 𝑥𝑑𝑦
𝑑𝑥− 𝑦 = 0
3. Form the differential equation corresponding to 𝑦2 − 2𝑎𝑦 + 𝑥2 = 𝑎2 by eliminating a.
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4. Form the differential equation representing the family of curves 𝑦 = 𝑎 𝑠𝑖𝑛 (𝑥 + 𝑏),
where 𝑎, 𝑏 are arbitrary constants.
5. Solve the differential equation.
𝑠𝑒𝑐2𝑥 𝑡𝑎𝑛 𝑦 𝑑𝑥 + 𝑠𝑒𝑐2𝑦 𝑡𝑎𝑛 𝑥 𝑑𝑦 = 0
6. Solve the differential equation.
𝑑𝑦
𝑑𝑥=
1+ 𝑦2
1+ 𝑥2
LEVEL II
Solve the following differential equations.
1. 𝑑𝑦
𝑑𝑥+
1
𝑥 . 𝑦 = 2𝑥2
2. (𝑥2 + 𝑥𝑦)𝑑𝑦 = (𝑥2 + 𝑦2)𝑑𝑥
3. 𝑑𝑦
𝑑𝑥+
4𝑥
𝑥2+1 𝑦 +
1
(𝑥2+1)2 = 0
4. 𝑠𝑖𝑛 𝑥 𝑑𝑦
𝑑𝑥+ 𝑐𝑜𝑠 𝑥. 𝑦 = 𝑐𝑜𝑠 𝑥. 𝑠𝑖𝑛2𝑥
5. 3𝑒𝑥 𝑡𝑎𝑛 𝑦 𝑑𝑥 + (1 − 𝑒2)𝑠𝑒𝑐2 𝑦 𝑑𝑦 = 0, 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑦 = 𝜋
4, 𝑤ℎ𝑒𝑛 𝑥 = 1.
6. (𝑥3 + 𝑦3)𝑑𝑦 − 𝑥2𝑦 𝑑𝑥 = 0
7. 𝑐𝑜𝑠2𝑥 𝑑𝑦
𝑑𝑥+ 𝑦 = 𝑡𝑎𝑛 𝑥
8. 𝑑𝑦
𝑑𝑥+ 𝑦 = 𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛 𝑥
9. 𝑥 𝑙𝑜𝑔 𝑥 𝑑𝑦
𝑑𝑥+ 𝑦 =
2
𝑥 𝑙𝑜𝑔 𝑥
LEVEL III
Solve the following differential equations
1. y – x 𝑑𝑦
𝑑𝑥 = a (𝑦2 + 𝑥2 𝑑𝑦
𝑑𝑥)
2. ( 1 + sin2x ) dy + ( 1 + y2 ) cos x dx = 0, given that x = 𝜋
2 , y = 0
3. ( x3 + x2 + x +1) 𝑑𝑦
𝑑𝑥 = 2x2 + x