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

1. \((x-z)=k e^{-\tan ^{-1} y}\)
2. \(\mathbf{Z x} \cdot \mathbf{e}^{\tan ^{-1} \mathbf{y}}=\left[\mathbf{e}^{\tan ^{-1} \mathbf{y}}\right]^{2}+k\)
3. \(\mathbf{x} \cdot \mathbf{e}^{\tan ^{-1} \mathbf{y}}=\tan ^{-1} \mathbf{y}+\mathbf{k}\)
4. \(\mathbf{x} \cdot \mathbf{e}^{2 \tan ^{-1} \mathbf{y}}=\mathbf{e}^{\tan ^{-1} \mathbf{y}}+k\)

Answer

(B) Z x e ^ ^ -1 y = [ e ^ ^ -1 y ]^ 2 +k

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