UPSC 2016 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Linear first-order · ODEs · asked 5× in 13 yrs · Read the full method →

Question

Solve: $\dfrac{dy}{dx}=\dfrac{1}{1+x^2}\left(e^{\tan^{-1}x}-y\right)$ .

Linear first-order ODE; integrating factor $\mu=e^{\arctan x}$ ; substitution $t=\arctan x$ to integrate $\int e^{2t}dt$ .

\frac{dy}{dx}+\frac{1}{1+x^2}\,y=\frac{e^{\tan^{-1}x}}{1+x^2}.

This is linear in $y$ with $P(x)=\dfrac{1}{1+x^2}$ , $Q(x)=\dfrac{e^{\tan^{-1}x}}{1+x^2}$ .

\int P\,dx=\int\frac{dx}{1+x^2}=\tan^{-1}x,\qquad \mu=e^{\tan^{-1}x}.

\frac{d}{dx}\!\left(y\,e^{\tan^{-1}x}\right)=e^{\tan^{-1}x}\cdot\frac{e^{\tan^{-1}x}}{1+x^2}=\frac{e^{2\tan^{-1}x}}{1+x^2}.

Substitute $t=\tan^{-1}x,\ dt=\dfrac{dx}{1+x^2}$ :

y\,e^{\tan^{-1}x}=\int e^{2t}\,dt=\frac{1}{2}e^{2t}+C=\frac{1}{2}e^{2\tan^{-1}x}+C.

\boxed{\;y=\frac{1}{2}e^{\tan^{-1}x}+C\,e^{-\tan^{-1}x}.\;}