First Order Differential Equation $(2x/y-y/(x^2+y^2))dx+(x/(x^2+y^2)-x^2/y^2)dy=0$

Question

Where can I start to solve this differential equation: $(2x/y-y/(x^2+y^2))dx+(x/(x^2+y^2)-x^2/y^2)dy=0$ . Differential Equation Image
My Solution

Answer 1

$$ \dfrac {2xdx}y-\dfrac {x^2}{y^2}dy +{xdy-ydx\over x^2+y^2}=0$$ The DE becomes: $$ \dfrac {2xdx}y-\dfrac {x^2}{y^2}dy +d \left(\arctan \left ( \dfrac y x \right) \right)=0$$

$$ \dfrac {dx^2}y+x^2d(\dfrac 1 y)+d \left(\arctan \left ( \dfrac y x \right) \right)=0$$ $$ d(\dfrac {x^2}y)+d \left(\arctan \left ( \dfrac y x \right) \right)=0$$ Integrate: $$ \dfrac {x^2}y+ \arctan \left ( \dfrac y x \right) =C$$

Answer 2

Hint

By defining $u=\tan^{-1} {x\over y}$ and $w={x^2\over y}$ we obtain $$du={dx\over y(1+{x^2\over y^2})}-{dy\over x(1+{y^2\over x^2})}={ydx-xdy\over x^2+y^2}$$ $$ dw={2xdx\over y}-{x^2dy\over y^2} $$

Answer 3

Sir, which part is wrong? Is -arctan(x/y)=arctan(y/x) ?

@Aryadeva

My Solution

Answer 4

$$\left(\dfrac {2x}y-\dfrac y{(x^2+y^2)}\right)dx+\left(\dfrac x{(x^2+y^2)}-\dfrac {x^2}{y^2}\right)dy=0$$ You can solve it with exactness. Since we have $$Mdx+Ndy=0$$ $$\partial_yM=\partial_xN$$ $$I=\int \dfrac {2x}y-\dfrac y{(x^2+y^2)}dx$$ $$I= \dfrac {x^2}y-\dfrac {dx/y}{((x/y)^2+1)}$$ $$I= \dfrac {x^2}y-\arctan (x/y)+C(y)$$ $$I= \dfrac {x^2}y+\arctan \left(\dfrac yx \right)+C_2(y)$$

On the other part we have: $$I= \int \dfrac x{(x^2+y^2)}-\dfrac {x^2}{y^2}dy$$ $$I=\arctan \left(\dfrac yx \right)+\dfrac {x^2}{y}+C(x)$$ So we have : $$C(x)=C_2(y) \implies C(x)=C$$ We can deduce the final answer as: $$\boxed {\arctan \left(\dfrac yx \right)+\dfrac {x^2}{y}=C}$$