Question

Find the singular solution of the differential equation $y=px+p-p^2$. Where $p=\frac{dy}{dx}$.

Answer 1

Given the differential equation,

$y=px+p-p^2$.......(1)

Now, differentiating both sides with respect to $x$ we get,

$p=p+(x+1-2p)\frac{dp}{dx}$ [ Since $\frac{dy}{dx}=p$]

or, $(x+1-2p)\frac{dp}{dx}=0$

or, $x+1=2p$ and $\frac{dp}{dx}=0$.

or, $p=\frac{x+1}{2}$ gives particular solution and $\frac{dp}{dx}=0$ leads to general solution.

Now putting $p=\frac{x+1}{2}$ in (1) we get,

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

or, $y=\frac{(x+1{)}^2}{4}$

or, $(x+1{)}^2=4y$.