y1

y2

...

yn

W[ y

, y

2

,..., y

n

]

y1

y2

...

yn

.

1

...

...

...

...

y(n 1)

y (n 1)

...

y(n 1)

1

2

n

For functions y1 (x), y2 (x), ..., yn (x)

continuous within the interval (a, b),

T 2.

Typical problems

1. Find the general solution

y e2x

sin x.

Solution. By n successive integrations, we get

y ¢¢¢ =

dy ¢¢

= e

2x

+sin x ,

y ¢¢ = ò (e

2x

+sin x)dx

=

1

e

2x

-cos x +C1 ,

dx

2

1

e

2x

cos x

1

e

2x

sin x

C1 x C2 ,

y y dx

2

C1 dx

4

1

2x

1

2x

C1

2

y

y dx

e

sin x C x C

dx

e

cos x

x

C

x C

,

4

8

2

1

2

2

3

2.

Solve

2

x

)

y (1

2xy .

Solution. The given equation contains derivatives of the dependent variable

y but does not contain y directly, then the substitution

y z(x) (see table 3.1).

Then y z and we obtain

x

2

) 2xz.

z (1

Separating the variables in the equation and integrating, we have

dz (1 x2 ) 2xz

,

dz

2x

dx ,

z

x 2

dx

1

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

z =C (1+x2 ) ,

dy = C (1+ x2 ) ,

1

1

dx

1

y C1 (1 x2 )dx C1 (x

x3

) C2

is general solution of the given equation.

3

3.

Solve

2

2yy

0.

(y )

Solution. An equation does not contain

x , directly, the substitution

y

p( y) ,

dp

y

p dy

p

2 2yp dp

0;

p( p 2y dp ) 0 ,

dy

dy

thereafter, the following two cases:

1)

p 0 , dy

= 0 , y =C ;

dx

2) p +2y dp = 0 , 1 dy + dp = 0 ,

1 ln

y

+ln

p

= ln C

,

dy

2

y

p

2

1

3

y p =C1

dy

ò

ydy = ò C1dx ,

2

y

, y

=C1 ,

2

C1 x C2 –

dx

3