JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \((x+y+2)^2 d x=d y, y(0)=-2\). Let the maximum and minimum values of the function \(y=y(x)\) in \(\left[0, \frac{\pi}{3}\right]\) be \(\alpha\) and \(\beta\), respectively. If \((3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in \mathbb{Z}\), then \(\gamma+\delta\) equals ....................
- A \(45\)
- B \(31\)
- C \(43\)
- D \(75\)
Answer & Solution
Correct Answer
(B) \(31\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{x}+\mathrm{y}+2)^2\) \(y(0)=-2\) Let \(\mathrm{x}+\mathrm{y}+2=\mathrm{v}\) \(1+\frac{d y}{d x}=\frac{d v}{d x}\) from (1) \(\frac{d v}{d x}=1+v^2\) \( \int \frac{d v}{1+v^2}=\int d x \) \( \tan ^{-1}(v)=x+C \)…
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