Let y = y(x) be the solution of the differential equation (x^2 + 4)^2 dy + (2x^3y + 8xy - 2) dx = 0.

Question

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

(1) \(\frac{\pi}{8}\)

(2) \(\frac{\pi}{16}\)

(3) \(2 \pi\)

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

Answer 1

Correct option is (4) \(\frac{\pi}{32}\)

\(\frac{d y}{d x}+y\left(\frac{2 x^{3}+8 x}{\left(x^{2}+4\right)^{2}}\right)=\frac{2}{\left(x^{2}+4\right)^{2}}\)

\(\frac{d y}{d x}+y\left(\frac{2 x}{x^{2}+4}\right)=\frac{2}{\left(x^{2}+4\right)^{2}}\)

\(\mathrm {I F}=e^{\int \frac{2 x}{x^{2}+4} d x}\)

\(\mathrm{IF}=\mathrm{x}^{2}+4\)

\(y \times\left(x^{2}+4\right)=\int \frac{2}{\left(x^{2}+4\right)^{2}} \times\left(x^{2}+4\right)\)

\(y\left(x^{2}+4\right)=2 \int \frac{d x}{x^{2}+2^{2}}\)

\(y\left(x^{2}+4\right)=\frac{2}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c\)

\(0=0+\mathrm{c}=\mathrm{c}=0\)

\(y\left(x^{2}+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)\)

y at \(x=2\)

\(y(4+4)=\tan ^{-1}(1)\)

\(y(2)=\frac{\pi}{32}\)