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(y/x)+y/x•SIN(y/x) → (1)

Answer 3

Transform:

Answer 4

(Y/x) → ø:

Answer 5

Y'=xø'+ø

Answer 6

(1) becomes:

Answer 7

Ø+xø'

Answer 8

=cscø+ø•SIN Ø → (2)

Answer 9

Now{øSINødø

Answer 10

=-{ø d(cosø)

Answer 11

=SINØ-Ø•COSØ

Answer 12

So that

Answer 13

So that (SINØ-Ø•COSØ)'

Answer 14

=ØSINØ → (3)

Answer 15

{dø/sinø

Answer 16

_={2dz/z^2 -1

Answer 17

Because sinø

Answer 18

=1/2i(z- 1/z),

Answer 19

Z_= VIS ^ upper ø

Answer 20

And dø

Answer 21

=dz/iz :

Answer 22

{2dz/z^2 -1

Answer 23

= ln z-1/z+1

Answer 24

And {dø/SINø

Answer 25

= ln{CIS ^ UPPER Ø-1/CIS ^ UPPER Ø+1}

Answer 26

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

Answer 27

= ln{TAN(Ø/2)}

Answer 28

That is,

Answer 29

(LnTAN ø/2)'

Answer 30

=CSCØ → (4)

Answer 31

(2),(3),(4) →

Answer 32

(2),(3),(4) →Ø+xØ'

Answer 33

=[(SINØ-Ø•COSØ)+LNTAN ø/2]'

Answer 34

In short

Answer 35

In short ø+xø'

Answer 36

G' → (xØ)'=G'

Answer 37

→(G-xø)'=0

Answer 38

G-xø=c ~ upper

Answer 39

→G= c ~ upper+xø

Answer 40

G= y +c ~ upper

Answer 41

Y/x=ø

Answer 42

& y= G -c ~ upper

Answer 43

G_= G(y,x)

Answer 44

The solution is

Answer 45

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

Answer 46

Therefore,