The solution of the differential equation d y d x + 2 y x 1+x^ 2 = 1 (1+x^ 2 )^ 2 is,
Mathematics · JEE Main · NTA Exams — Differential Equations
The solution of the differential equation \(\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\) is,
- \(y\left[1+x^{2}\right]=c+\tan ^{-1} x\)
- \(\frac{y}{1+x^{2}}=c+\tan ^{-1} x\)
- \(y \log \left[1+x^{2}\right]=c+\tan ^{-1} x\)
- \(y\left(1+x^{2}\right)=c+\sin ^{-1} x\)
Answer
(A) y [1+x^ 2 ]=c+ ^ -1 x
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