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

Let \(y=y(x)\) be the solution of the differential equation \(\left(x^2+4\right)^2 d y+\left(2 x^3 y+8 x y-2\right) d x=0 \text {. If } y(0)=0 \text {, }\) then \(y(2)\) is equal to

  1. A \(\frac{\pi}{8}\)
  2. B  \(\frac{\pi}{16}\)
  3. C \(2 \pi\)
  4. D \(\frac{\pi}{32}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\pi}{32}\)

Step-by-step Solution

Detailed explanation

\( \frac{d y}{d x}+y\left(\frac{2 x^3+8 x}{\left(x^2+4\right)^2}\right)=\frac{2}{\left(x^2+4\right)^2} \) \( \frac{d y}{d x}+y\left(\frac{2 x}{x^2+4}\right)=\frac{2}{\left(x^2+4\right)^2}\) \( \text { IF }=e^{\int \frac{2 x}{x^2+4} d x} \) \( \text { IF }=x^2+4 \)…
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