T1. Typical problems

xydx (1 x 2 ) ln ydy 0 .

Solution. Division by

y(1 x2 ) 0 gives variables separable:

x

dx

ln y

dy 0,

1 x2

y

integrating this, we get

1 ln

1 ln 2 y C .

1 x2

2

2

x 1 .

The given equation has the other singular solution

2.

Find

the

general

solution

and

also

the

particular solution of

y

y

through the point

y(0)

e .

1 x2

Solution. Separating the variables, we obtain

dy

=

y

,

ò dy

= ò

dx

,

ln

y

arcsin x C .

dx

1-x

2

y

1-x

2

Substituting of (0, e) in the general solution gives

ln e arcsin 0 C C 1 .

Hence, ln

y

arcsin x 1 is a particular solution for (0, e).

3. Find the general integral of

y

¢

x(1+ y2 )

= y(1+ x2 ) .

is the general integral of y

¢

x(1+ y2 )

= y(1+x2 ) .

4. Solve the equation

y

y2

.

xy x2

Solution. Since the equation is homogeneous, write

y xu(x) , dy

x du

u .

dx

dx

x

du

+u =

u2 x2

,

x

du

u

u 2

,

x

du

=

u

,

dx

x2u -x2

dx

u

1

dx

u

-1

u 1

du dx ,

u ln

u

ln

x

ln

C

,

u ln

Cxu

.

u

x

y

Replacing u by its equal

, we have

x

y x ln

Cy

.

5. Find the general solution of

y

x 2y 3

.

2x 2

Solution.

Finding the

determinant

=

1

2

4 0 . Since

0 ,

2

0

transforming this to the homogeneous equation substituting

x t ,

y v .

By (3.4), we get 2 3 0,

then 1,

1.

Thus

2 2 0,

dy

d (v 1)

dv .

x t 1,

y v 1,

dx

d(t 1)

dt

Substituting the values of t and v, we obtain

dv

t 1 2v 2 3

, or

dv

t 2v

.

dt

2t 2 2

dt

2t

Since this equation is homogeneous, write

v ut,

u u(t),

v

u t u.

Then

t

2ut

1 2u

1

du

1

dt

u t u

; u t

u

2

; u t

2

; 2

dt

t

;

2du

t .

2t

198

8. Solve

y

2x

2

y y

0 .

x2 1

y

2x

1 1.

y2

x2 1

y

This form suggests the substitution

1 ,

dy

1

dz .

z = y-1 . Then

y =

=-

dx

z2

Substituting, we get

z

dx

dz

2x

1

dz

2x 1

1

0 or

z 1 .

x2 1 z

z 2

dx

x2 1

z 2 dx

dz

2x

z

0

,

dz

2x

dx ,

dz

d(x2 1)

,

dx

x

2

z

x

2

z

x

2

1

1

1

ln | z | ln | C | ln(x2 1) ,

z C(x2 1).

Since С is an unknown function of x , namely C =C(x) , then

z =C(x)(x2 +1) ,

dz

dC

(x 2 1) 2xC(x) .

dx

dx

Substituting this in

dz

2x

z

0 , we get

dx

x2

dC

1

2x

(x2

1) C(x)2x

C(x)(x2 1) 1,

dx

x2

therefore

1

dC

1

.

Then

dx

x2 1

dx

C(x)

arctg x C

( C const ).

1

1

1

Finally,

2

yex + y2 =C –

2

1

y (arctg x C )(x2

1)

1

is the general solution of the given equation.

y 0.

The given equation has another singular solution

9. Solve the equation

yex dx +( y +ex )dy = 0.

Solution. From §1.5, it appears that an equation

is exact (3.11) if

Mdx + Ndy = 0

M

N .

y

x

In our case, we have

дM

дN

M (x, y) ye x ,

N (x, y) y e x ;

e x ,

e x .

дy

дx

Therefore M

N , and the equation is exact. Consequently,

y

x

дu

M (x, y) ye

x

,

дx

дu N (x, y) y e x .

дy

Integrating of the first equation with respect to x gives

u ye x dx ye x

( y).

( 3.17 )

Differentiating with respect to y gives

дu

x

y : дy e

( y) , or

e

x

x

,

¢

( y) = y ,

y 2

C .

( y) y e

j

2

1

Thus,

u ye x

y 2

C .

x) ,

Y y y (X

y

М

where Х,

Y are the set

of points in the

1

coordinate plane of the tangent. As the given

tangent intercepts the line Y = 1 at point 2х,

О

х 2х

therefore

dy

Х

¢

Fig. 3.1

1- y

(2x -x) , або

x dx =1- y .

= y

This is the variables separable differential equation. Thereafter

dy

=- dx , ln | y -1|=-ln | x | +ln C ,

y -1= C .

y -1

x

x

where acceleration a dv . In our case

F = dv . Simultaneous by the equation

dt

dt

F k

t

occurs, where

k from 4 k 10

is k = 20. Comparing the right parts of

v

50

F = dv

and F = 20

t

, we then write our differential equation of motion in

v

dt