solve: (x+y•sin^2(y/x)) dx=x^2 SIN(y/x) dy

Question

(x+y•SIN^2(y/x)) dx=x^2 SIN(y/x) dy

Answer 1

By rearrangement:

Answer 2

xy'=csc(t/x)+y/x •SIN(y/x → (1)

Answer 3

Transform,

Answer 4

( y/x)

Answer 5

→ Ø :

Answer 6

Y'=xø'+ ø

Answer 7

(1) becomes:

Answer 8

ø + xø'

Answer 9

= cscø+ ø• SINØ → (2)

Answer 10

Now { Ø SINØ dø

Answer 11

=-{ød(cosø)

Answer 12

= SINØ- Ø•COSØ

Answer 13

So that,

Answer 14

(SINØ-Ø•COSØ)'=ØSINØ

Answer 15

{dø/SINØ_={2dz/z^2 -1

Answer 16

Because

Answer 17

SINØ=1/2i(z-1/2)

Answer 18

Z_=CISØ

Answer 19

And dø=dz/iz:

Answer 20

{2dz/z^2 -1

Answer 21

= ln z-1/z+1

Answer 22

And {dø/SINØ

Answer 23

= ln{cis ^ upper Ø -1/cis ^ upperØ+1}

Answer 24

= ln{cis^ upper (ø/2)-cis ^ upper(-ø/2)/cis ^ upper(ø/2)+cis ^ upper(-ø/2)}

Answer 25

= ln{TAN(ø/2)}

Answer 26

That is:

Answer 27

( LUTANø/2)'

Answer 28

=CSCØ → (4)

Answer 29

(2),(3),(4)→

Answer 30

(2),(3),(4) →Ø+xØ'

Answer 31

=[(SINØ-Ø•COSØ)+ LUTAN Ø/2]'

Answer 32

In short,

Answer 33

Ø+xØ'=G'

Answer 34

→(xø)'

Answer 35

= G' →

Answer 36

→ (G-xø)'=0

Answer 37

G-xø=c ~ upper

Answer 38

→ G=c ~ upper +xø

Answer 39

G= y+ c ~ upper

Answer 40

Y/x=ø

Answer 41

& y= G-c ~ upper,

Answer 42

G_= G(y,x)

Answer 43

The solution is,

Answer 44

Y=SIN(y/x)-(y/x)•COS(y/x)+ Lu{ C•TAN( y/2x)}

Answer 45

Therefore,