Question

The solution of $\frac{dy}{dx}+\frac{y}{x}=\frac{1}{\sqrt{1+x^2}}$ is

• A. $y=\frac{1+x^2}{x}+\frac{c}{x}$
• B. $y=\frac{\sqrt{1+x^2}}{x}+\frac{c}{x}$
• C. $y=\frac{x}{\sqrt{1+x^2}}+cx$
• D. none of these

Answer 1

Answer: (B)

Given $\frac{dy}{dx}+\frac{y}{x}=\frac{1}{\sqrt{1+x^2}}$
It is linear differential equation with $I.F.=e^{\int \frac{1}{x}dx}=x$
$\therefore$ solution is, $y\cdot x=\int \frac{x}{\sqrt{1+x^2}}dx=\frac{1}{2}\int \frac{2x}{\sqrt{1+x^2}}dx$
$\Rightarrow y\cdot x=\sqrt{1+x^2}+c$
$\Rightarrow y=\frac{\sqrt{1+x^2}}{x}+\frac{c}{x}$