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

જે વિક્લ સમીકરણ \(\left(x^4+2 x^3+3 x^2+2 x+2\right) \mathrm{d} y-\left(2 x^2+2 x+3\right) \mathrm{d} x=0\) નો ઉકલ \(y=y(x)\) એ \(y(-1)=-\frac{\pi}{4}\) નું સમાધાન કરે, તો \(y(0)=\) ...........

  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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