The solution of the differential equation (1+y^ 2 )+ (x-e^ ^ -1 y ) d y d x =0 is
Mathematics · JEE Main · NTA Exams — Differential Equations
The solution of the differential equation
\(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\) is
\(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\) is
- \((x-2)=c e^{-2 \tan ^{-1} y}\)
- \(2 x \mathrm{e}^{\tan ^{-1} y}=\mathrm{e}^{2 \tan ^{-1} y}+c\)
- \(x e^{\tan ^{-1} y}=\tan ^{-1} y+c\)
- \(\mathrm{xe}^{2 \tan ^{-1} \mathrm{y}}=\mathrm{e}^{\tan ^{-1} \mathrm{y}}+\mathrm{c}\)
Answer
(B) 2 x e ^ ^ -1 y = e ^ 2 ^ -1 y +c
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