Question

Verify $y=cx+\frac{1}{x}$ is a solution of the differential equations $y\frac{dy}{dx}=x{\left(\frac{dy}{dx}\right)}^2+1$, where c is arbitrary constant.

Answer 1

since the given equation is

$y\frac{dy}{dx}=x{\left(\frac{dy}{dx}\right)}^2+1let,\frac{dy}{dx}=p\therefore yp=x{p}^2+1⟶\left(1\right)$

differentiating w.r.t x

$\frac{dy}{dx}.p+y\frac{dp}{dx}=p^2+2xp\frac{dp}{dx}\Rightarrow y\frac{dp}{dx}-2xp\frac{dp}{dx}=p^2-\frac{dy}{dx}.p\Rightarrow \frac{dp}{dx}(y-2xp)=0(since\frac{dy}{dx}=p)\therefore \frac{dp}{dx}=0ory-2xp=0\Rightarrow p=constantorp=\frac{y}{2x}\because p=\frac{y}{2x}\Rightarrow \frac{dy}{dx}=\frac{y}{2x}\Rightarrow \frac{dy}{y}=\frac{dx}{2x}integratingbothside\int \frac{dy}{y}=\int \frac{dx}{2x}\Rightarrow logy=log2x+logC\Rightarrow logy=log2x.C(sinceloga+logb=loga.b)\therefore y=2xC$

hence $y=Cx+\frac{1}{x}$ is not a solution of the differential equation