JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}, y(1)=0\). Then \(\mathrm{y}(0)\) is
- A \(\frac{1}{4}\left(\mathrm{e}^{\pi / 2}-1\right)\)
- B \(\frac{1}{2}\left(1-\mathrm{e}^{\pi / 2}\right)\)
- C \(\frac{1}{4}\left(1-\mathrm{e}^{\pi / 2}\right)\)
- D \(\frac{1}{2}\left(\mathrm{e}^{\pi / 2}-1\right)\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\left(1-\mathrm{e}^{\pi / 2}\right)\)
Step-by-step Solution
Detailed explanation
\( \frac{d y}{d x}+\frac{y}{1+x^2}=\frac{e^{\tan ^{-1} x}}{1+\mathrm{x}^2} \) \( \text { I.F. }=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\tan ^{-1} \mathrm{x}} \)…
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