Mathematics - Differential Equation Question with Solution | TestHub

The given equation is $y=2px+y^2{p}^3$ .......(i)

Solving for $x,x=\frac{y}{2p}-\frac{1}{2}{y}^2{p}^3$

Differentiating w.r.t y

$\frac{dx}{dy}=\frac{1}{p}=\frac{1}{2p}-\frac{y}{2p^2}\cdot \frac{dp}{dy}-y{p}^2-y^{2}p\cdot \frac{dp}{dy}$

or $2p=p-y\frac{dp}{dy}-2y{p}^3-2y^2{p}^2\frac{dp}{dy}$

or $p\left(!+2y{p}^2\right)+y\frac{dp}{dy}\left(1+2y{p}^2\right)=0$

or $\left(1+2y{p}^2\right)\left(p+y\frac{dp}{dy}\right)=0$

Neglecting the first factor which does not involve $\frac{dp}{dy}$, we have

$p+y\frac{dp}{dy}=0$

$\Rightarrow \frac{dp}{p}+\frac{dy}{y}=0$

Integrating $\log p+\log y=\log c$

or $\log py=\log c$

$\Rightarrow py=c$ .......(ii)

Eliminating $p$ between (i) and (ii)

$y=2x\cdot \frac{c}{y}+y^2\cdot \frac{c^3}{y^3}$

or $y=\frac{2cx}{y}+\frac{c^3}{y}$

or $y^2=2cx+c^3$ which is the required solution