JEE Mains · Maths · STD 12 - 9. differential equations
If the solution \(y=y(x)\) of the differential equation \(\left(\mathrm{x}^4+2 \mathrm{x}^3+3 \mathrm{x}^2+2 \mathrm{x}+2\right) \mathrm{dy}-\left(2 \mathrm{x}^2+2 \mathrm{x}+3\right) \mathrm{dx}=0\) satisfies \(y(-1)=-\frac{\pi}{4}\), then \(y(0)\) is equal to :
- A \(-\frac{\pi}{12}\)
- B \(0\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{2}\)
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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