Solve the following differential equation $$ \left(x^2+1\right) \frac{d y}{d x}+4 x y=\frac{1}{x^2+1} $$
Solve the following differential equation
$$
\left(x^2+1\right) \frac{d y}{d x}+4 x y=\frac{1}{x^2+1}
$$
- $y\left(x^2-1\right)^2=x+c$
- $y\left(x^2+1\right)^2=x+c$
- $y\left(x^2+1\right)^2=x^2+c$
- $y\left(x^2-1\right)^2=x^2+c$
Solution
Given, differential equation
$
\begin{aligned}
\left(x^2+1\right) \frac{d y}{d x}+4 x y & =\frac{1}{x^2+1} \\
\frac{d y}{d x}+\frac{4 x y}{x^2+1} & =\frac{1}{\left(x^2+1\right)^2} \\
\mathrm{IF}=e^{\int \frac{4 x}{x^2+1} d x} & =e^{2 \log \left(x^2+1\right)} \\
& =\left(x^2+1\right)^2
\end{aligned}
$
Solution of the differential equation is
$
\begin{aligned}
& y\left(x^2+1\right)^2=\int d x+c \\
& y\left(x^2+1\right)^2=x+c
\end{aligned}
$
Asked in: AP EAMCET 2021 (25 Aug Shift 1)
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