Let y=y(x) be the solution curve of the differential equation [ y d y~d x+2 x y=x^3 y, y(1…

Question

Let $y=y(x)$ be the solution curve of the differential equation $$\sec y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y, y(1)=0$$. Then $$y(\sqrt{3})$$ is equal to:

Accepted Answer

$$\begin{aligned}
& \sec y \frac{d y}{d x}+2 x \sin y=x^3 \cos y \
& \Rightarrow \sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3
\end{aligned}$$

Let $$z=\tan y$$

$$\begin{aligned}
& \frac{d z}{d x}=\sec ^2 y \frac{d y}{d x} \
& \Rightarrow \frac{d z}{d x}+2 x z=x^3
\end{aligned}$$

$$\begin{aligned}
& \text { I.F. }=e^{x^2} \
& \Rightarrow z . e^{x^2}=\int e^{x^2} \cdot x^3 d x+c
\end{aligned}$$

$$\Rightarrow \tan y \cdot e^{x^2}=\frac{1}{2}\left(x^2 e^{x^2}-e^{x^2}\right)+c$$

$$\Rightarrow \tan (0) \cdot e=\frac{1}{2}(1 \cdot e-e)+c$$

$$\Rightarrow c=0$$

$$\Rightarrow \tan y=\frac{x^2-1}{2}$$

$$f(x)=\tan ^{-1}\left(\frac{x^2-1}{2}\right) \Rightarrow f(\sqrt{3})=\frac{\pi}{4}$$

Solutions · 1