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.