The particular solution of the different equation $$ \frac{d x}{d y}=\frac{\sin y(1+y \cot y)}{x \log…
The particular solution of the different equation
$$
\frac{d x}{d y}=\frac{\sin y(1+y \cot y)}{x \log \left(x^2 e\right)}, y(1)=0
$$
- $y \sin y=x^2 \log x$
- $y^2 \sin y=\log x$
- $y=\left(\frac{e^2}{\sin e}\right)(x-1)$
- $y=e^2 \sec x$
Solution
$\begin{aligned} \text { } & \frac{d x}{d y}=\frac{\sin y(1+y \cot y)}{x \log \left(\mathrm{x}^2 \mathrm{e}\right)}, y(1)=0 \\ \Rightarrow & \int x \log \left(x^2 e\right) d x=\int(\sin y+y \cos y) d y \\ & \text { Let } x^2 e=t \\ \Rightarrow & x d x=\frac{1}{2 e} d t \\ \Rightarrow & \frac{1}{2 e} \int \log t d t=-\cos y+y \sin y-\int \frac{d y}{d y} \cdot \sin y d y+C \\ \Rightarrow & \frac{1}{2 e}[t \log t-t]=y \sin y+c\end{aligned}$
$
\begin{aligned}
& \Rightarrow \quad \frac{x^2 e}{2 e}\left(\log x^2 e-1\right)=y \sin y+c \\
& \Rightarrow \quad \frac{x^2}{2}\left(\log x^2+\log e-1\right)=y \sin y+C \\
& \Rightarrow \quad x^2 \log x=y \sin y+C \quad\left\{\because \log _{\mathrm{e}^{\mathrm{e}}}=1\right\} \\
& \text { When } x=1, y=0 \\
& \Rightarrow \quad 1 \log 1=0+C \\
& \Rightarrow C=0
\end{aligned}
$
Hence $x^2 \log x=y \sin y$
Asked in: AP EAMCET 2023 (19 May Shift 1)
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