Question

Solve the differential equation x^3 p^2 +x^2 yp+a^3=0 and also obtain its singular solution, if
it exists.

Answer 1

$x^3 p^2 +x^2 yp+a^3=0$ ---(1) is the given differential equation, where $p=dy/dx$

$\implies y=-xp-a^3/(x^2p)$

Differentiating with respect to x, we get;

$p=-p-xdp/dx+a^3/x^2p^2(dp/dx)+2a^3/x^3p$

$\implies 2p+xdp/dx=2a^3/px^3+a^3/x^3p^2(dp/dx)$

$\implies 2p(1-a^3/x^3p^2)+xdp/dx(1-a^3/x^3p^2)=0$

$\implies 2p+xdp/dx=0$

$\implies\int dp/p=-2\int dx/x$

$\implies px^2=c \implies p=c/x^2$ ---(2)

Substituting (2) in (1), we get;

$c^2+xyc+a^3x=0$ which is a quadratic equation in c;

Equating it's discriminant to 0, we get;

$x^2y^2-4a^3x=0$

$\implies x(xy^2-4a^3)=0$

$\implies x=0 andxy^2-4a^3=0$

Both these satisfy (1).

Thus, $x=0$ and $xy^2-4a^3=0$ are the singular solutions of the given differential equation.