Solve the differential equation: $(x^2-y^2)dx+2xydy=0$.

Question

Given $(x^2-y^2)dx+2xydy=0$

My solution-

Divide the differential equation by $dx$

$\Rightarrow x^2-y^2+2xy\frac{dy}{dx}=0$

$\Rightarrow 2xy\frac{dy}{dx}=y^2-x^2$

Divide both sides by $2xy$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}[\frac{y}{x}-\frac{x}{y}]$

This is a homogenous differential equation. Substitute $y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}[v-\frac{1}{v}]$

$\Rightarrow x\frac{dv}{dx}=-\frac{v^2+1}{2v}$

$\Rightarrow -\frac{2v}{v^2+1}dv=\frac{dx}{x}$

Integrating both sides

$\Rightarrow -\log|v^2+1|=\log x+\log c$

$\Rightarrow -\log|\frac{y^2}{x^2}+1|=\log xc$

$\Rightarrow -\log|\frac{x^2+y^2}{x^2}|=\log xc$

$\Rightarrow \frac{x^2}{x^2+y^2}= xc$

$\Rightarrow x= c(x^2+y^2)$

$\Rightarrow y=\pm \sqrt{xc-x^2}$

Kindly review my solution and let me know if there are other methods of solving such problems.

Answer 1

Your answer is correct. Here is another way to solve the DE: $$(x^2-y^2)dx+2xydy=0$$ Divide by $x^2dx$: $$1+\dfrac {2xyy'-y^2}{x^2}=0$$ $$1+\left (\dfrac {y^2}{x}\right)'=0$$ Integrate: $$x+\dfrac {y^2}{x}=C$$

Answer 2

Another approach let's call $u=x^2+y^2$.

Then $xu'=2x^2+2xyy'=2x^2+(y^2-x^2)=x^2+y^2=u$

Which solves to $u=cx\iff x^2+y^2=cx$

Answer 3

Your solution it is correct. Answering your second question, here is another slightly different solution.

If we can write the ODE as a Bernoulli equation: $y'+a(x)y=b(x)y^{\alpha}$ for some real $\alpha$ so the substitution $v=y^{1-\alpha}$ transforms the Bernoulli equation into a linear equation $\frac{1}{1-\alpha}v'+a(x)v=b(x)$ which can be solved using integrating factor.

We can re-write the ODE as $$(x^2-y^2)dx+2xydy=0$$ $$(x^2-y^2)+2xy\frac{dy}{dx}=0$$ $$2xy\frac{dy}{dx}-y^2=-x^2$$ $$\frac{dy}{dx}-\frac{1}{2x}y=-\frac{x}{2}y^{-1}$$ Consider $a(x)=\frac{-1}{2x}$ and $b(x)=\frac{-x}{2}$ so the equation is Bernoulli and we can make the substitution $v=y^{1-(-1)}=y^{2}$ and then we can write the ODE as $$\frac{1}{2}\frac{dv}{dx}-\frac{1}{2x}v=-\frac{x}{2}$$ $$\frac{dv}{dx}-\frac{1}{x}v=-x.$$

The equation it is linear and using your integrating factor $\mu(x)=e^{\int -\frac{1}{x}dx}=\frac{1}{x}$ we can transforms the linear equation into a separable equation.

$$\frac{d}{dx}\left(\mu(x)v\right)=-x\mu(x)$$ Integrating, $$\frac{v}{x}=\int-1dx=-x+c$$ Thus, $$v=-x^2+cx$$ Substitution back $$y^2=-x^2+cx$$ Therefore, the general solution is given by $$ y=\pm\sqrt{-x^2+cx}$$ as you said.