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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
4.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
A differential equation is an equation which relates an unknown function of a single variable with one or more of its derivatives.
Example:
The order of a differential equation is the highest derivative involved in the equation, and the degree is the power of the highest derivative in the equation.
Example :
a.
dydx
= 1+ x1+ y2
is a DE of order 1 and degree 1.
b.x2 dydx
+ y sin x is a DE of order 1 and degree 1.
c.
d2 ydx2
+4dydx
−2 y=x2
is a DE of order 2 and degree 1.
d.xydydx
−xy2 = x is a DE of order 1 and degree 1.
e.y2+1 = x( dydx )
2
is a DE of order 1 and degree 2
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Independent variable
Derivative of y wrt x
Dependent variable
B5001- Engineering Mathematics DIFFERENTIAL EQUATION
f. is a DE of order 2 and degree 2.
Example 1
State the dependent variable, the independent variable, the order and degree for each differential equation.
i.
dsdt
= st ii.
dmdn
−2mn = 0
iii.( dydx )
2
= 4+xx3
iv.t 2dsdt
+ sin t = 0
v.(t 2−1 ) dy
dt = yt
vi.xdydx
+xy = y2+1
vii.xd2 ydx 2
+4dydx
+2 xy = 0viii.
(u3+1 ) dvdu
−uv2 = u
ix.( d2 ydx2 )
2
+3dydx
+2 y = 0x.
xd2 ydx 2
+4dydx
+2 xy = 0
Answer
Dependent VariableIndependent
variableOrder Degree
i. s t 1 1
ii. m n 1 1
iii. y x 1 2
iv. s t 1 1
v. y t 1 1
vi. y x 1 1
vii. y x 2 1
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
viii. v u 1 1
ix. y x 2 2
x. y x 2 1
4.2 Formation of Differential Equations
If ; Differential Equation is formed when the arbitrary constant A
is eliminated from this equation.
Example 2
i. If
Differentiate with respect to x,
dydx
= 20x
dydx
−20 x = 0
ii. If y = x+Ax2 ------Differentiate with respect to x, gives,
dydx
= 1+2 Ax -----
From , A =
y−xx2
Subtituting A =
y−xx2
into
Then,
dydx
= 1+2( y−xx2 )x
= 1+2( y−xx )
¿ 1+2
yx−2
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Differential equation.
B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Multiply both sides with x.
So, xdydx
= x+2 y−2x
xdydx
= 2 y−x
iii. If y = Ax2−Bx+x -----
Differentiate with respect to x, gives,
dydx
= 2 Ax−B+1 -----
Differentiate
dydx with respect to x.
d2 ydx2
= 2 A -----
From , A =
12d2 ydx 2 -----
Substituting into to get B.
dydx
= 2(12d2 ydx2 )x−B+1
= x( d2 y
dx 2 )−B+1
B = x ( d2 ydx2 )−dydx +1
-----
Then, substitute and into .
y = ( 1
2d2 ydx2 ) x2−(x d2 y
dx 2−dydx
+1) x+ x
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
= 12x2 d
2 ydx 2
−x2 d2 ydx2
+x dydx
−x+x
= − 12x2 d
2 ydx 2
+x dydx
EXERCISE 1
Form differential equation of the following:
i. y=x3+Ax2 iv. y=4 Bx+A
ii. y=Ax2+7 x v. y=A sin 2 x+B kos x
iii. y=Dx2+Ex vi. y=Ce−x+2De x
Answer
i.x2 dydx
=3x 4−2( y−x3 )iv
xdydx
=2 y−7 x
ii.y=x dy
dx− x
2
2 ( d2 ydx2 )
v.
d2 ydx2
=0
iii.
d2 ydx2
+4 y=0vi
d2 ydx2
− y=0
4.3 SOLUTIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS (DE)
We already know that DE is one that contains differential coefficient.
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Example: i. - 1st order DE
ii. - 2nd order DE
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
4.3.1 - Solve By Direct Integration; i.e.
Example 3: Find the general solution of the DE
Answer
Example 4: Find the particular solution of de , given the boundary
conditions .
Answer
Hence
Substituting the boundary conditions;
The particular solution,
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
4.3.2 Solve By Variable Separable;
4.3.2.1 - rearranged to give and then solve by direct
integration; i.e.
Example 5: Find the general solution of de .
Answer
i.e.
Example 6: Find the particular solution of de given that y = 1 when
Answer
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Substituting y=1 and ;
4.3.2.2 - rearranged to give and then solve
by direct integration
Example 7: Solve
Answer
Separating the variables gives: (3 -2y)dy = (2x3 – 1)dx
Example 8: The current in an electric circuit containing resistance R and inductance L
in series with a constant voltage source E is given by the de .
Solve the equation and find i in terms of time t given that when t = 0 and i
= 0.
Answer
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
When t = 0, i= 0;
Example 9: Solve the Des:
a. d.
b. e.
c.
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
4.3.3 Homogeneneous First Order DEs
An equation of the form , where P and Q are function of both x and y of the
same degree – said to be homogenous in y and x.
Example;
f(x,y) Homogeneous degree
1 x2 + 3xy + y2 yes 2
2 yes 1
3 yes 2
4 noy – 1
x - 2
Procedure to solve DE of the form
i.Rearrange into the form
ii.Make the substitution y =vx, from , by the product rule.
iii. Substitute for both y and in the equation . Simplify, by
cancelling, and on equation result in which the variables are separable.
iv. Separate the variable and solve as direct integrating.
v.Substitute
to solve in terms of the original variables.
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Example 10: Solve the DE , given x = 1 when y = 2.Answer
i.Rearranging;
ii.
iii. Substitute for both y and gives:
Example 11: Find the particular solution of DE; given the boundary
conditions that y = 4 when x = 1.
Answer
i.Rearranging;
ii.
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
iii.
4.3.4 Linear First Order DEs
If P = P(x) and Q = Q(x) are functions of x only, then is called a linear
differential equation order 1. We can solve these linear DEs using an integrating
factor. For linear DEs of order 1, the integrating factor is: e∫Pdx
The solution for the DE is given by multiplying y by the integrating factor (on the left)
and multiplying Q by the integrating factor (on the right) and integrating the right side
with respect to x, as follows:
Example 12: Solve for
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Answer
Now for the integrating factor:
IF=
For the left hand side of the formula
we have ye∫Pdx = yx-3
For the right hand of the formula, Q = 7 and the IF = x-3, so:
Applying the outer integral:
Now, applying the whole formula;
we have ;
Multiplying throughout by x3 gives:
Example 13: Solve
Answer
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Here,
Determine
IF =
Now
Apply the formula:
The integral needs a simple substitution: u = sin x, du = cos x dx
Divide throughout by sin x:
Example 14: Solve
Answer
Dividing throughout by dx to get the equation in the required form, we get:
In this example, P(x) = 3 and Q(x) = e-3x.
Now e∫Pdx = e∫3dx = e3x
and
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Using , we have:
ye3x = x + K
or we could write it as:
Example 15: Solve 2(y - 4x2)dx + x dy = 0
Answer
We need to get the equation in the form of a linear DE of order 1.
Expand the bracket and divide throughout by dx:
Rearrange:
Divide throughout by x:
Here,
IF
Now
Applying the formula:
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
gives:
Divide throughout by x2:
Example 16: Solve
Answer
Divide throughout by x:
Here,
Now
Applying the formula: gives
This requires integration by parts, with
So
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Multiplying throughout by x4 gives:
Example 17: Solve , x ¿0 , subject to the initial condition y = 2
when x = 0
Answer
The differential equation can be expressed in the proper form by adding
2y to both sides:
for x ¿0
We have P(x) = 2 and Q(x) =
An integrating factor is given by IF = e for x ¿0
Next, use the first-order linear differential equation theorem, where IF =
e2 xand Q(x) = , to find y:
y =
= for x ¿0
To find C, insert y = 2 when x = 0, to obtain C = 1
Thus y = , x ¿0
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Example 18: Solve with y = 1 when x = 0
Answer
P(x) = ; Q(x) = sin x
IF = e = e
y =
=
=
=
Since y = 1 when x = 0,
1 = implies C = 0
y =
Example 19:
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
a. Solve the DE , given the boundary conditions y=1 when θ = 0.
[Ans: y = ( + 1) sec ]θ θ
b. Solve the DE . [ ans: ]
c. Consider a simple electric circuit with the resistance of 3 an inductance of 2H. If a
battery gives a constant voltage of 24V and the switch is closed when t = 0, the
current, I(t), after t seconds is given by
i) Obtain I(t) [ans: ]
ii) Determine the difference in the amount of current flowing through the circuit
from the fourth to eight seconds. Give your answer to 3 d.p.
[ans: 0.05 A]
iii) If the current is allowed to flow through the circuit for a very long period of
time, estimate I(t).
[ans: ]
4.4 SECOND ORDER DIFFERENTIAL EQUATION
The general form of the second order differential equation with constant coefficients is
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where a, b, c are constants with a > 0 and Q(x) is a function of x only.
4.4.1 Homogeneous Equation
In this section, most of our examples are homogeneous 2nd order linear DEs (that is, with
Q(x) = 0):
, where a, b, c are constants.
Method of Solution
The equation am2 + bm + c = 0 is called the Auxiliary Equation (A.E.) (or
Characteristic Equation)
The general solution of the differential equation depends on the solution of the A.E. To
find the general solution, we must determine the roots of the A.E. The roots of the A.E.
are given by the well-known quadratic formula:
Summary:
If Differential Equation: ay'' + by' + c = 0 and Associated auxiliary equation is:
am2 + bm + c = 0
Nature of roots Condition General Solution
4.4.1. Real and distinct roots,
m1, m2
b2 − 4ac > 0 y = Aem1x + Bem
2x
4.4.2. Real and equal roots, m b2 − 4ac = 0 y = emx(A + Bx)
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
4.4.3. Complex roots
m1 = α + jω
m1 = α − jω
b2 − 4ac < 0 y = eαx(A cos ωx + B sin ωx)
Example 20
The current i flowing through a circuit is given by the equation ,
Solve for the current i at time t > 0.
Answer
The auxiliary equation arising from the given differential equations is:
A.E.:
So and
We have 2 distinct real roots, so we need to use the first solution from the table above
(y = Aem1x + Bem
2x), but we use i instead of y, and t instead of x.
So
We could have written this as:
Since we have 2 constants of integration. We would be able to find these constants if
we were given some initial conditions.
Example 21
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Solve the following equation in which s is the displacement of a object at time t.
, given that s = 1,
when t = 0
(That is, the object's position is 1 unit and its velocity is 3 units at the beginning of the
motion.)
Answer
The auxiliary equation for our differential equation is:
A.E.
In this case, we have: (repeated root or real equal roots)
We need to use the second form from the table above (y = emx(A + Bx)), and once again
use the correct variables (t and i, instead of x and y).
So .
Now to find the values of the constants:
So we can write
So
The graph of our solution is as follows:
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
Example 22
Solve the equation
Answer
This time the auxiliary equation is:
Solving for m, we find that the solutions are a complex conjugate pair:
The solution for our DE, using the 3rd type from the table above:
y = eαx(A cos ωx + B sin ωx)
we get:
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Example 23
In a RCL series circuit, R = 10 Ω, C = 0.02 F, L = 1 H and the voltage source is E = 100
V. Solve for the current i(t) in the circuit given that at time t = 0, the current in the circuit
is zero and the charge in the capacitor is 0.1 C.
[Note: Damping and the Natural Response in RLC Circuits
Consider a series RLC circuit (one that has a resistor, an inductor and a capacitor) with
a constant driving electro-motive force (emf) E. The current equation for the circuit is
Differentiating, we have
; This is a second order linear homogeneous equation.]
Answer
AE : ,
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B5001- Engineering Mathematics DIFFERENTIAL EQUATION
The factors are:
So, ; ; (This means at t = 0, i = A = 0 in this
case.)
Then , We need to find the value of B.
Differentiating gives:
At t = 0,
Returning to equation
Now, at time t = 0, ; So , so B = 19.
Therefore,
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