Question

Solve: $(x^3+3x{y}^2)dx+(y^3+3x^{2}y)dy=0$

Answer 1

$\Rightarrow \frac{dy}{dx}=-\frac{(x^3+3x{y}^2)}{(y^3+3x^{2}y)}=-\frac{3x{y}^2(\frac{x^3}{3x{y}^2}+1)}{3x^{2}y(\frac{y^3}{3x^{2}y}+1)}=-\frac{y(\frac{x^2}{3y^2}+1)}{x(\frac{y^2}{3x^2}+1)}$

$\Rightarrow \frac{dy}{dx}=f\left(\frac{y}{x}\right)$

$\Rightarrow$ the given differential equation is a homogeneous equation.

The solution of the given differential equation is :

Put $y=vx$

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

$v+x\frac{dv}{dx}=-\frac{vx(\frac{x^2}{3(vx{)}^2}+1)}{x\left(\frac{(vx{)}^2}{3x^2}\right)+1}=-v\frac{(\frac{1}{3(v{)}^2}+1)}{(\frac{(v{)}^2}{3}+1)}=-\frac{1+3(v{)}^2}{3+(v{)}^2}\times \frac{1}{v}$

$=-\frac{1+3(v{)}^2}{3v+(v{)}^3}$

$\Rightarrow x\frac{dv}{dx}=-\frac{1+3(v{)}^2}{3v+(v{)}^3}-v=-\frac{1+3(v{)}^2+3(v{)}^2+(v{)}^4}{3v+(v{)}^3}=\frac{1+6(v{)}^2+(v{)}^4}{3v+(v{)}^3}$

$\Rightarrow \frac{3v+(v{)}^3}{1+6(v{)}^2+(v{)}^4}dv=-\frac{dx}{x}$

Integrating both the sides we get:

$\Rightarrow \int \frac{3v+(v{)}^3}{1+6(v{)}^2+(v{)}^4}dv=-\int \frac{dx}{x}+c$

As we know that, $\frac{d}{dv}(1+6(v{)}^2+\left(v{)}^4\right)=12v+4v^3=4(3v+v^3)$

$\Rightarrow \frac{ln|1+6(v{)}^2+(v{)}^4}{4}+ln|x|=ln|c|$

Resubstituting the value of $y=vx$ we get

$\Rightarrow \frac{ln|1+6(\frac{y}{x}{)}^2+(\frac{y}{x}{)}^4}{4}+ln|x|=ln|c|$

$\Rightarrow y^4+6x^2{y}^2+x^{3}4=C$

Ans :$\Rightarrow y^4+6x^2{y}^2+x^{3}4=C$