(a)OR(b)

Solve the differential equation $\displaystyle 2(y+3)-x y \frac{d y}{d x}=0$; given $\displaystyle y(1)=-2$.

Given differential equation can be written as \[\begin{aligned} & \frac{y}{y+3} d y=\frac{2}{x} d x \\ & \Rightarrow \int\left(1-\frac{3}{y+3}\right) d y=2 \int \frac{1}{x} d x \\ & \Rightarrow y-3 \log |y+3|=2 \log |x|+C \\ & y=-2, \text { when } x=1 \Rightarrow C=-2 \end{aligned} \] Hence, the required particular solution is \[\Rightarrow y-3 \log |y+3|=2 \log |x|-2 \]Given differential equation can be written as $\displaystyle \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{4 x^{2}}{1+x^{2}}$, which is linear in y . \[\text { I.F. }=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2} \] The solution is given by \[y\left(1+x^{2}\right)=\int 4 x^{2} d x \] $\displaystyle \Rightarrow y\left(1+x^{2}\right)=\frac{4}{3} x^{3}+C$ or $\displaystyle y=\frac{4 x^{3}}{3\left(1+x^{2}\right)}+C \frac{1}{\left(1+x^{2}\right)}$, which is the required general solution