if `x^2=2y^2logy` then prove `(x^2+y^2)(dy)/(dx)-xy=0`

if x^(2)=2y^(2)log y then prove (x^(2)+y^(2))(dy)/(dx)-xy=0

Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation: x^2 = 2y^2 (logy) : (x^2 + y^2)((dy)/(dx)) - xy = 0

For each of the exercises given below, verify that the given function ( implicit or explicit ) is a solution of the corresponding differential equation. x^(2)=2y^(2)logy:(x^(2)+y^(2))(dy)/(dx)-xy=0

Solve :- (x^2+y^2)dx-2xy dy=0

If y="sin"(logx), then prove that (x^2d^2y)/(dx^2)+x(dy)/(dx)+y=0

(dy)/(dx)=xy+2y+x+2

If xy log(x + y) = 1 , then prove that (dy)/(dx) = -(y(x^(2)y + x + y))/(x(xy^(2) + x + y)) .

if y=x+(1)/(x) prove that x^(2)(dy)/(dx)-xy+2=0

If y=sin(logx) , prove that x^2(d^2y)/(dx^2)+x(dy)/(dx)+y=0 .