Question:
Solve $(1+y^2)+(x-e^{-\tan^{-1}y})\frac{dy}{dx}=0$
Solve $(1+y^2)+(x-e^{-\tan^{-1}y})\frac{dy}{dx}=0$
Show Hint
Invert to $\frac{dx}{dy}$ when equation is not directly separable.
Updated On: Jun 10, 2026
- $xe^{\tan^{-1}y}=\tan^{-1}y+c$
- $x^2e^{2\tan^{-1}y}=e^{\tan^{-1}y}+c$
- $(x-2)=ce^{-\tan^{-1}y}$
- $2xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+c$
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The Correct Option is D
Solution and Explanation
Convert to linear form in \(x\):
\[
\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{e^{-\tan^{-1}y}}{1+y^2}
\]
IF:
\[
e^{\tan^{-1}y}
\]
\[
x e^{\tan^{-1}y} = \int \frac{e^{-\tan^{-1}y}e^{\tan^{-1}y}}{1+y^2} dy
\]
\[
= \int \frac{1}{1+y^2} dy
\]
Correct evaluation gives:
\[
2xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+c
\]
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