Mathematics - Differential Equation Question with Solution | TestHub

Given,

$\left(3y^2-5x^2\right)ydx+2x\left(x^2-y^2\right)dy=0$

$\Rightarrow \frac{dy}{dx}=\frac{y\left(5x^2-3y^2\right)}{2x\left(x^2-y^2\right)}$

This is homogeneous differential equation

Let $y=tx\Rightarrow \frac{dy}{dx}=t+x\frac{dt}{dx}$

$t+x\frac{dt}{dx}=\frac{\left(tx\right)\left(5x^2-3t^2{x}^2\right)}{2x\left(x^2-t^2{x}^2\right)}=\frac{t\left(5-3t^2\right)}{2\left(1-t^2\right)}$

$\Rightarrow x\frac{dt}{dx}=\frac{5t-3t^3}{2-2t^2}-t=\frac{3t-t^3}{2-2t^2}$

$\Rightarrow \frac{2\left(1-t^2\right)}{3t-t^3}dt=\frac{1}{x}dx$

Taking integration on both sides we get,

$\int \frac{2\left(1-t^2\right)}{3t-t^3}dt=\int \frac{1}{x}dx......\left(1\right)$

Now let $3t-t^3=z\Rightarrow 3\left(1-t^2\right)dt=dz$

$\Rightarrow \left(1-t^2\right)dt=\frac{1}{3}dz$

Substitute in equation $\left(1\right)$ we get,

$\frac{2}{3}\int \frac{dz}{z}=\int \frac{1}{x}dx$

$\Rightarrow \frac{2}{3}\ln \left(\mathrm{z}\right)=\ln x+\ln c$

$\Rightarrow z^{\frac{2}{3}}=cx\Rightarrow z=c^{\frac{3}{2}}{x}^{\frac{3}{2}}$

$\Rightarrow 3t-t^3=c^{\frac{3}{2}}{x}^{\frac{3}{2}}$

Now put $t=\frac{y}{x}$

$\Rightarrow \frac{3y}{x}-\frac{y^3}{x^3}=c^{\frac{3}{2}}{x}^{\frac{3}{2}}$

Given $y\left(1\right)=1$ when $x=1,y=1$

 $\Rightarrow c^{\frac{3}{2}}=2$

So, equation becomes $3\frac{y}{x}-\frac{y^3}{x^3}=2x^{\frac{3}{2}}$

$\Rightarrow 3x^{2}y-y^3=2x^{\frac{9}{2}}......\left(2\right)$

Put $x=2$ in equation $\left(2\right)$,

$12y\left(2\right)-{\left(y\left(2\right)\right)}^3=2(2{)}^{\frac{9}{2}}=32\sqrt{2}$

$\therefore \left|{\left(y\left(2\right)\right)}^3-12y\left(2\right)\right|=32\sqrt{2}$