Question

$\frac{dy}{dx}+\frac{4x}{x^2+1}y+\frac{1}{{\left(x^2+1\right)}^2}=0$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+\frac{4x}{x^2+1}y+\frac{1}{{\left(x^2+1\right)}^2}=0 \\ \Rightarrow \frac{dy}{dx}+\frac{4x}{x^2+1}y=-\frac{1}{{\left(x^2+1\right)}^2}.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dy}{dx}+Py=Q \\ \mathrm{where} \\ P=\frac{4x}{x^2+1} \\ Q=-\frac{1}{{\left(x^2+1\right)}^2} \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdx} \\ =e^{2\int \frac{2x}{x^2+1}dx} \\ =e^{2\log \left|x^2+1\right|} \\ ={\left(x^2+1\right)}^2 \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}{\left(x^2+1\right)}^2,\mathrm{we}\mathrm{get} \\ {\left(x^2+1\right)}^2\left(\frac{dy}{dx}+\frac{4x}{x^2+1}y\right)={\left(x^2+1\right)}^2\left[-\frac{1}{{\left(x^2+1\right)}^2}\right] \\ \Rightarrow {\left(x^2+1\right)}^2\frac{dy}{dx}+4x\left(x^2+1\right)y=-1 \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ {\left(x^2+1\right)}^{2}y=-\int dx+C \\ \Rightarrow {\left(x^2+1\right)}^{2}y=-x+C \\ \mathrm{Hence},{\left(x^2+1\right)}^{2}y=-x+C\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$