Question

Prove that $x^2-y^2=c(x^2+y^2)$is the general solution of differential equation($x^3-3xy^2)dx=(y^3-3x^2y)dy$,where $c$ is parameter.

Answer 1

$(x^3-3xy^2)dx=(y^3-3x^2y)dy$

$⇒\frac{dy}{dx}=\frac{x^3-3xy^2}{y^3-3x^2y}...(1)$

This is a homogenous equation.To simplify it,we need to make the substitution as:

$y=vx$

$⇒\frac{d}{dx}(y)=\frac{d}{dx}(vx)$

$⇒\frac{dy}{dx}=v+x\frac{dv}{dx}$

Substituting the values of $y$ and $\frac{dy}{dx}$ in equation(1),we get:

$v+x\frac{dv}{dx}=\frac{x^3-3x(vx)^2}{(vx)^3-3x^2(vx)}$

$⇒v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}$

$⇒x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v-v}$

$⇒x\frac{dv}{dx}=\frac{1-3v^2-v(v^3-3v)}{v^3-3v}$

$⇒x\frac{dv}{dx}=\frac{1-v^4}{v^3-3v}$

$⇒(\frac{v^3-3v}{1-v^4})dv=\frac{dx}{x}$

Integrating both sides,we get:

$\int(\frac{v^3-3v}{1-v^4})dv=logx+logC'...(2)$

Now,$\int(\frac{v^3-3v}{1-v^4})dv=\int\frac{v^3dv}{1-v^4}-3\int\frac{vdv}{1-v^4}$

$⇒\int(\frac{v^3-3v}{1-v^4})dv=I_1-3I_2,where\space I_1=\int\frac{v^3dv}{1-v^4}\space and \space I_2=\int\frac{vdv}{1-v^4}....(3)$

Let $1-v^4=t.$

$∴\frac{d}{dv}(1-v^4)=\frac{dt}{dv}$

$⇒-4v^3=\frac{dt}{dv}$

$⇒v^3dv=-\frac{dt}{4}$

⇒Now,$I_1=\int-\frac{dt}{4t}=-\frac{1}{4}logt=-\frac{1}{4}log(1-v^4)$

And,$I_2=\int\frac{vdv}{1-v^4}=\int\frac{vdv}{1-(v^2)^2}$

Let $v^2=p$

$∴\frac{d}{dv}(v^2)=\frac{dp}{dv}$

$⇒2v=\frac{dp}{dv}$

$⇒vdv=\frac{dp}{2}$

$⇒I_2=\frac{1}{2}\int\frac{dp}{1-p^2}=\frac{1}{2\times2}log|\frac{1+p}{1-p}|=\frac{1}{4}log|\frac{1+v^2}{1-v^2}|$

Substituting the values of I1 and I2 in equation(3),we get:

$\int(\frac{v^3-3v}{1-v^4})dv=-\frac{1}{4}log(1-v^4)-\frac{3}{4}log|\frac{1-v^2}{1+v^2}|$

Therefore,equation(2),becomes:

$\frac{1}{4}log(1-v^4)-\frac{3}{4}log|\frac{1+v^2}{1-v^2}|logx+logC'$

$⇒-\frac{1}{4}log[(1-v^4)(\frac{1+v^2}{1-v^2})]=logC'x$

$⇒\frac{(1+v^2)^4}{(1-v^2)^2}=(C'x)-4$

$⇒\frac{(1+\frac{y^2}{x^2})^4}{(1-\frac{y^2}{x^2})^2}=\frac{1}{C'4x^4}$

$⇒\frac{(x^2+y^2)^4}{x^4(x^2-y^2)^2}=\frac{1}{CX'4x^4}$

$⇒(x^2-y^2)^2=C'^4(x^2+y^2)^4$

$⇒(x^2-y^2)=C'^2(x^2+y^2)^2$

$⇒x^2-y^2=C(x^2+y^2)^2,where C=C'^2$

Hence,the given result is proved.