solve: Y(x^2+y)dx+x(x^2-2y)dy=0 , Y=2 for x=1

Question

Y(x^2+y)dx+x(x^2-2y)dy=0 , Y=2 for x=1

Answer 1

The given can be written ad Pdx+Qdy=0

Answer 2

It is not exact because Py-Qx=4y-2x^2

Answer 3

→I.F=1/x^2

Answer 4

Then P/x^2 dx+Q/x^2 dy=0 is exact

Answer 5

And du_=P/x^2 dx+Q/x^2 dy=0

Answer 6

→ u(x,y)=c

Answer 7

æu/æy=x-2y/x

Answer 8

→u=w(x)+xy-y^2/x (1)

Answer 9

æu/æx=w'(x)+y+y^2/x^2_=y+y^2/x^2

Answer 10

→w(x)=x (2)

Answer 11

(1),(2)→ yx-y^2/x+x=c

Answer 12

With y(1)=2

Answer 13

Therefore y^2-x^2•y-x(x+1)=0

Answer 14

Solving in respect of Y with Determinant

Answer 15

D=x^4+4x•(x+1)=(x^2+2)^2

Answer 16

The solutions are: y={-1 upper, x^2+1 lower