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JEE Mains · Maths · STD 12 - 9. differential equations

यदि अवकल समीकरण \(\left(\mathrm{x}^4+2 \mathrm{x}^3+3 \mathrm{x}^2+2 \mathrm{x}+2\right) \mathrm{dy}-\left(2 \mathrm{x}^2+2 \mathrm{x}+3\right) \mathrm{dx}=0\) का हल \(y=y(x)\), \(y(-1)=-\frac{\pi}{4}\) को संतुष्ट करता है, तो \(y(0)\) = ...........

  1. A  \(-\frac{\pi}{12}\)
  2. B \(0\)
  3. C \(\frac{\pi}{4}\)
  4. D \(\frac{\pi}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{\pi}{4}\)

Step-by-step Solution

Detailed explanation

\( \int d y=\int \frac{\left(2 x^2+2 x+3\right)}{x^4+2 x^3+3 x^2+2 x+2} d x \) \( y=\int \frac{\left(2 x^2+2 x+3\right)}{\left(x^2+1\right)\left(x^2+2 x+2\right)} d x \) \( y=\int \frac{d x}{x^2+2 x+2}+\int \frac{d x}{x^2+1} \) \( y=\tan ^{-1}(x+1)+\tan ^{-1} x+C \)…
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