Question

Find the solution of $(2ax+x^2)\frac{dy}{dx}=a^2+2ax$

• A. $y+k=\frac{a}{2}[\log x+3\log (x+2a)]$
• B. $y+k=\frac{a}{2}[\log x-3\log (x+2a)]$
• C. $y+k=\frac{a}{2}[\log x+4\log (x+2a)]$
• D. $y+k=\frac{a}{2}[\log x-4\log (x+2a)]$

Answer 1

The correct option is A $y+k=\frac{a}{2}[\log x+3\log (x+2a)]$

Given, $(2ax+x^2)\frac{dy}{dx}=a^2+2ax$
$dy=\frac{a(a+2x)}{x(x+2a)}dx$
Split into P.F.s.
$dy=\frac{a}{2}[\frac{1}{x}+\frac{3}{x+2a}]dx.$
Integrating we get,
$\int dy=\int \frac{a}{2}[\frac{1}{x}+\frac{3}{x+2a}]dx.$
$y+k=\frac{a}{2}[\log x+3\log (x+2a)]$