The Integration factor the differential equation $\left(1+x^{2}\right) d t=\left(\tan ^{-1} x-t\right) d x$
The Integration factor the differential equation $\left(1+x^{2}\right) d t=\left(\tan ^{-1} x-t\right) d x$
- $-e^{\frac{\left(\tan ^{-1} x\right)^{2}}{2}}$
- $-e^{\tan ^{-1} x}$
- $e^{\frac{\left(\tan ^{-1} x\right)^{2}}{2}}$
- $e^{\tan ^{-1} x}$
Solution
$\begin{aligned}
\frac{\mathrm{dt}}{\mathrm{dx}} &=\frac{\tan ^{-1}-\mathrm{t}}{1+\mathrm{x}^{2}} \\
\therefore \frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{1+\mathrm{x}^{2}} &=\frac{\tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}} \\
\text { I.F. }=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}} &=\mathrm{e}^{\tan ^{-1} \mathrm{x}}
\end{aligned}$
Asked in: MHT CET 2020 (16 Oct Shift 1)
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