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Math 201 Solutions to Assignment 1

1. Solve the initial value problem:

x2dx + 2ydy = 0, y(0) = 2.

Solution:

x2dx + 2ydy = 0 , y(0) = 2

2ydy =

x2dx

y2 = 13

x3 + C

y =

C 13

x3

Notice that y is not defined for some values of x. Applying the initialcondition gives us that y(0) =

C = 2, so C = 4. Then

y =

4 1

3x3 .

2. Find the general solution to the equation:

xdy

dx+ 3(y + x2) =

sin x

x.

Solution:

The integrating factor is given by:

(x) =3

x

and after integration we find that (x) = x3. Multiply the standardform of the equation by to obtain

(x3y) = 3x4 + x sin x

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Integrating both sides we get

x3y = 3/5x5 +

x sin xdx.

integrating the integral above by parts and dividing the equation by x3

we obtain

y = 3/5x2 cos x/x2 + sin x/x3 + C/x3.

3. Solve the initial value problems:

(yexy

1/y)dx + (xexy + x/y2)dy = 0, y(1) = 1,

Solution:

Using the test for exactness one verifies that the equation is exact.Then

F/x = yexy 1/yand after integration

F(x, y) = exy x/y + g(y) (1)Differentiating with respect to y we get

F/y = xexy + x/y2 + g(y)

On the other hand we need that

F/y = xexy + x/y2

and therefore g(y) = 0. Thus we can choose g(y) = 0 because anyconstant can be absorbed in the right hand side constant of the solutionwhich we easily obtain from (1) to be:

exy x

y = const = C

Since we want y(1) = 1 it follows that

F(1, 1) = e 1 = C

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and then the solution of the initial-value problem (IVP) is given by

exy xy

= e 1.

(etx + 1)dt + (et 1)dx = 0, x(1) = 1.

Solution: Again we can verify that this is an exact equation. Then

F/t = etx + 1

and integrating we obtain

F(t, x) = xet + t + g(x)

Differentiating with respect to y

F/x = et + g(x)

On the other hand

F/x = e

t

1 g(x) = 1

or g = x

F(t, x) = xet + t x = const = C

is a solution. We need that x(1) = 1 or

F(1, 1) = e = C

which gives the unique solution to the IVP

x =e t

et 1

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4. Find the general solution to the equations:

dy

dx=

y(ln y ln x + 1)x

.

Solution:

The equation can be put in the form

dy

dx=

y

x

ln

y

x+ 1

.

This is a homogeneous equation and we need to substitute u = y/x.Then the equation in terms of u reads

xdu

dx+ u = u(ln u + 1).

This is a separable equation and the separated form is

du

u ln u=

dx

x.

After one integration we obtain

lnln |u| = ln |x|+ c1and taking the exponents of both sides

ln |u| = cx.This yields

y = xecx.

dy

dx y = e2xy3

Solution: We first divide by y3 and obtain

y3dy

dx y2 = e2x.

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Note that while dividing by y we presume that y = 0 but y = 0 is asolution to the initial equation which will not be a solution to the newequation. Therefore, we have to keep it in mind. We further substitutey2 = u and after some simple calculus (du/dx = 2y3dy/dx) obtain

du

dx+ 2u = 2e2x.

This is a linear equation and the integrating factor is (x) = e2x. Mul-tiplying the equation with it we obtain

d

dx(e2xu) = 2e4x

and integrating once, and substituting back u = y2

y2 = 12

e2x + ce2x.

So the solutions are y2 = 12

e2x + ce2x and y = 0.

dy

dx= (2x + y

1)2,

Solution:

The equation is of the type: G(ax + by) with a = 2, b = 1. Wesubstitute v = 2x + y to obtain:

dv

dx= 2 +

dy

dx

and therefore the equation becomes

dv

dx= (v

1)2 + 2

which is a separable equation. Then

dv

(v 1)2 + 2 = dx

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which after one integration yields

12

arctanv 1

2= x + c1

orv =

2tan(

2x + c), c =

2c1

Substituting v = 2x + y we get

y = 1 2x +

2tan(

2x + c)

dy

d+

y

= 4y2,

Solution:

The equation is of a Bernoulli type with n = 2. Therefore we substi-tute v = y3 which yields

dy

d= 1/3y2

dv

d

The equation transforms into

1/3y2dv

d+

y

= 4y2

or (substituting y3 = v)

dv

d+

3

v = 12

which is a linear equation. (x) = 3 and multiplying by it we get

dd [3v] = 124

which after one integration gives

v = 12/52 + c/3

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or

y = 12/52/3

+ c1/

dy

dx=

x y 1x + y + 5

Solution:

Some elementary algebra (as shown in p. 73 of the textbook, 8thedition) shows that the substitution x = u 2, y = v 3 reduces theequation to the homegeneous one:

dv

du=

u vu + v

Therefore we substitute v/u = w which yields

dv

udu v

u2=

dw

du

Thendw

duu = w + 1 w

1 + w

After some elementary calculations we obtain

1 + w

1 2w w2dw =du

u

Integrating once both sides we get

1

2

dp

2p = ln |u|+ c

where p = (w + 1)2. After computing the integral we get

1/2 ln |(w + 1)2 2| = ln |u|+ c

or after taking the exponent of both sides

(w + 1)2 2 = Au2.

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After substituting back w, u and v and some elementary algebra we

obtain (y + 3)2 + 2(x + 2)(y + 3) (x + 2)2 = Aor

y2 x2 + 2xy + 10y + 2x = C

5. Solve the initial value problem:

dy

dx=

x

y+

y

x

, y(1) = 4.

Solution:

The equation is obviously homogeneous and therefore we substitute

v =y

x

which yieldsdy

dx= x

dv

dx+ v

After substitution in the equation we get

xdv

dx

=1

vwhich is a separable equation. After integrating it we obtain the fol-lowing solution

1/2v2 = ln |x|+ cor after substituting v = y/x

y2

x2= 2 ln |x|+ c

To obtain the solution of the IVP we impose that y2(1) = 16 which

yields c = 16.

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