(a) Find the particular solution of the differential equation dy/dx= y cot 2 x, given that y ((π)/4)=2.

$$\begin{aligned} & \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\mathrm{y} \cot 2 x \Rightarrow \int \frac{\mathrm{d} \mathrm{y}}{\mathrm{y}}=\int \cot 2 x \mathrm{d} x \\ & \Rightarrow \log |\mathrm{y}|=\frac{1}{2} \log |\sin 2 x|+\log c \\ & \mathrm{y}=c . \sqrt{\sin 2 x} \\ & \text { when } \mathrm{y}\left(\frac{\pi}{4}\right)=2, \text { gives } c=2 \\ & \therefore \mathrm{y}=2 \sqrt{\sin 2 x} \text { is the required Particular solution of given D.E. } \end{aligned} $$ $\displaystyle \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\mathrm{e}^{\frac{\mathrm{y}}{x}}+\frac{\mathrm{y}}{x}=f\left(\frac{\mathrm{y}}{x}\right)$ so, its a homogeneous differential equation Let $\displaystyle \mathrm{y}=v x \Rightarrow \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=v+x \frac{\mathrm{d} v}{\mathrm{d} x}$ Now, $\displaystyle v+x \frac{\mathrm{d} v}{\mathrm{d} x}=\mathrm{e}^{v}+v$ $$\begin{equation*} \Rightarrow \int \mathrm{e}^{-v} \mathrm{d} v=\int \frac{1}{x} \mathrm{d} x \tag{1} \end{equation*} $$ $\displaystyle \Rightarrow-\mathrm{e}^{-v}=\log |x|+c \Rightarrow-\mathrm{e}^{\frac{-\mathrm{y}}{x}}=\log |x|+c$. Now, $\displaystyle x=1, \mathrm{y}=1$, gives $\displaystyle c=-\mathrm{e}^{-1}$ Thus, $\displaystyle \log |x|+\mathrm{e}^{\frac{-\mathrm{y}}{x}}=\mathrm{e}^{-1}$