Hi all, I've attempted the question part b) after rearranging the 1st expression to : $$v(x,y)\, dx - u(x,y) \, dy = 0.$$
After this I tried using the method of differentiating each expression and trying to calculate if they were exact. It was very lengthy so I am unable to post on here but it seems there is an easier way to go about this. Any help is appreciated.
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The equation can be rewritten as $$\frac{1}{(x^2+y^2)^2}\left(2xdx+2ydy\right)y-\frac{2y^2}{(x^2+y^2)^2}dy+\frac{y^2-x^2}{(x^2+y^2)^2}dy+dy=0$$Thus$$y\frac{1}{(x^2+y^2)^2}d(x^2+y^2)+\left(-\frac{1}{x^2+y^2}\right)d(y)+dy=0$$which can be put up as $$y\left\{d\left(-\frac{1}{x^2+y^2}\right)\right\}+\left(-\frac{1}{x^2+y^2}\right)d(y)+dy=0$$Using the product rule of differentiation we have$$d\left\{\frac{-y}{x^2+y^2}\right\}+dy=0$$ So,integrating we get$$-\frac{y}{x^2+y^2}+y=c$$
$$(-v)dx+(u)dy=\frac{2xy}{(x^2+y^2)^2}dx+\left(1+\frac{y^2-x^2}{(x^2+y^2)^2} \right)dy=0$$ $$\frac{\partial (-v)}{\partial y}= \frac{\partial }{\partial y}\left(\frac{2xy}{(x^2+y^2)^2} \right)= \frac{2x}{(x^2+y^2)^2}-\frac{8xy^2}{(x^2+y^2)^3} =\frac{2x(x^2-3y^2)}{(x^2+y^2)^3}$$
$$\frac{\partial u}{\partial x}= \frac{\partial}{\partial x}\left(1+\frac{y^2-x^2}{(x^2+y^2)^2} \right)=\frac{-2x}{(x^2+y^2)^2}-\frac{4x(y^2-x^2)}{(x^2+y^2)^3} = \frac{2x(x^2-3y^2)}{(x^2+y^2)^3}$$ $$\frac{\partial (-v)}{\partial y}= \frac{\partial u}{\partial x}\quad\text{Thus the ODE is exact}$$
Solving the ODE : $$\int(-v)dx=\int\frac{2xy}{(x^2+y^2)^2}dx=\frac{-y}{(x^2+y^2)}+f(y)$$ $$\int(u)dx=\int\left(1+\frac{y^2-x^2}{(x^2+y^2)^2}\right)dx=\frac{-y}{(x^2+y^2)}+y+g(x)$$ $$\frac{-y}{(x^2+y^2)}+f(y)=\frac{-y}{(x^2+y^2)}+y+g(x) \quad\implies\quad f(y)=y\text{ and } g(x)=0$$ The ODE is transformed into : $\quad d\left(\frac{-y}{(x^2+y^2)}+y\right)=0$. Integrating leads to the solution of the ODE : $$\frac{-y}{(x^2+y^2)}+y=C$$