The solution of differential equation (x 2 –1) dy dx + 2 xy = 1 x^ 2 -1 is
Mathematics · JEE Main · NTA Exams — Differential Equations
The solution of differential equation
(x2 –1) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2 xy = \(\frac{1}{x^{2}-1}\) is
(x2 –1) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 2 xy = \(\frac{1}{x^{2}-1}\) is
- \(y\left(x^{2}-1\right)=\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|+C\)
- \(y\left(x^{2}-1\right)=\frac{5}{2} \log \left|\frac{x-1}{x+1}\right|+C\)
- none of these
Answer
(A) y (x^ 2 -1 )= 1 2 | x-1 x+1 |+C
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