Solving $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{xy+{y}^{2}}{{x}^{2}+{y}^{2}}$.

Question

Solve $$\frac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{xy+{y}^{2}}{{x}^{2}+{y}^{2}}.$$

I have tried to solve this question by assuming $y/x$ to be $v$, but I am stuck on the integral $$\int\frac{{v}^{2}+1}{{v}^{3}-1},$$ please help me with this integral or suggest a method to directly solve the differential equation.

Answer 1

After factoring, this becomes easier: $$\dfrac{dy}{dx} =\dfrac{xy+y^2}{x^2+y^2}=\dfrac{y\left(1+\frac yx\right)}{x\left(1+\left(\frac yx\right)^2\right)}$$ Substitute $u=\frac yx$ so that $y=ux$ and $\frac{dy}{dx}=u+x\frac{du}{dx}$, which implies $$u+x\dfrac{du}{dx}=\dfrac{u(1+u)}{1+u^2}\implies \dfrac{du}{dx} =\dfrac{u^2(1-u)}{x(1+u^2)}\implies\frac{u^2+1}{u^2(1-u)}\,du=\frac{1}{x}\,dx$$ This new DE is separable, so \begin{align*}\int\frac{u^2+1}{u^2(1-u)}\,du&=\int\frac{1}{x}\,dx\\ \int\left(\frac{2}{1-u}+\frac{1}{u}+\frac{1}{u^2}\right)\,du&=\int\frac{1}{x}\,dx\\ -2\ln|1-u|+\ln|u|-\frac{1}{u}&=\ln|x|+C_1\\ 2\ln|1-u|+\frac{1}{u}+\ln|x|-\ln|u|&=C_2\\ 2\ln\left|1-\frac yx\right|+\frac{x}{y}+\ln|x|-\ln\left|\frac yx\right|&=C_2 \end{align*}

As confirmation, here’s a slope field with some numerical solutions:

which matches up nicely with plots of the solution we found:

Answer 2

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{xy+{y}^{2}}{{x}^{2}+{y}^{2}}.$$ You made a mistake somewhere. Substitute $y=tx$ gives $$t'x+t=\dfrac{t^2+t}{1+t^2}$$ $$\dfrac {(t^2+1)}{t^2(1-t)}dt=\dfrac {dx}{x}$$ The integral that you need to evaluate is: $$I=\int \dfrac {(t^2+1)}{t^2(1-t)}dt$$ $$I=\int \dfrac {dt}{1-t}+\int \dfrac {dt}{t^2(1-t)}$$ The last integral can be evaluated by substitution.

Substitute $u=\dfrac 1 t$ $$I_2=-\int \dfrac {u}{u-1}du=-u+C-\int \dfrac {du}{u-1}$$