Question

Solve the following differential equation:
$x^2\frac{dy}{dx}=y^2+2xy$
Given that : $y=1$, when $x=1$.

Answer 1

$x^2\left(\frac{dy}{dx}\right)=y^2+2xy$

Rearranging , $\Rightarrow \left(\frac{dy}{dx}\right)=\frac{y^2}{x^2}+\frac{2xy}{x^2}={\left(\frac{y}{x}\right)}^2+2\left(\frac{y}{x}\right)$ ....(1)

Since the differential equation is of the form $\frac{dy}{dx}=f\left(y/x\right)$

we can use separation of variables theory

Let $y=kx$ substitute $y=kx$ in the equation ...(i)

$\Rightarrow \frac{dy}{dx}=k^2+2k$ ...(2)

and, since $y=kx\Rightarrow \frac{dy}{dx}=[k+x\left(\frac{dk}{dx}\right)]$ ....(3)

equating (2) & (3)

$k^2+2k=k+x\frac{dk}{dx}\Rightarrow x\frac{dk}{dx}=k^2+k\Rightarrow \frac{dk}{k^2+k}=\frac{dx}{x}$ ...(4)

$\Rightarrow \frac{dk}{k(k+1)}=(\frac{A}{k}+\frac{B}{k+1})dk=\left(\frac{A(k+2)+B\left(k\right)}{k\left(kH\right)}\right)dk=\left(\frac{A+Ak+Bk}{k\left(kH\right)}\right)dk$

$\Rightarrow A=1,A+B=0$ (equating coefficients on LHS & RHS)

$\Rightarrow A=1,B=-1$

$\Rightarrow \int \frac{dk}{k(k+1)}=\int \frac{dk}{k}-\int \frac{dk}{k+1}=\frac{dx}{x}\Rightarrow \log \left(k\right)-\log (k+1)=\log x+\log c$

$\Rightarrow \log \left(\frac{k}{k+1}\right)=\log \left(cx\right)\Rightarrow \frac{k}{k+1}=cx\Rightarrow \frac{\left[\frac{\left(y\right)}{\left(x\right)}\right]}{[\left(\frac{y}{x}\right)+1]}=cx$

when $x=1,y=1\Rightarrow \frac{1}{(1+1)}=1\left(c\right)\Rightarrow c=\frac{1}{2}$

$\Rightarrow y=(y+x)\frac{x}{2}$