(a) Find the particular solution of the differential equation dy/dx-2 x y =3 x^2 e^x^2; y (0)=5

Given differential equation is a linear order differential equation with:

$$P=-$\displaystyle 2$ \mathrm{x}, Q=$\displaystyle 3$ \mathrm{x}^{$\displaystyle 2$} \mathrm{e}^{\mathrm{x}^{$\displaystyle 2$}}

$$$$ \text { Integrating Factor }=\mathbf{e}^{\int-$\displaystyle 2$ \mathrm{xdx}}=\mathbf{e}^{-\mathrm{x}^{$\displaystyle 2$}}

$$The general solution is: $\displaystyle \mathrm{y} \cdot \mathrm{e}^{-\mathrm{x}^{2}}=\int \mathrm{e}^{-\mathrm{x}^{2}} \cdot 3 \mathrm{x}^{2} \mathrm{e}^{\mathrm{x}^{2}} d \mathrm{x}+C \Rightarrow \mathrm{y} \cdot \mathrm{e}^{-\mathrm{x}^{2}}=\mathrm{x}^{3}+C$

Putting $\displaystyle \mathrm{x}=0, \mathrm{y}=5$, we get, $\displaystyle C=5$

∴ The Particular solution is: $\displaystyle \mathbf{y} \cdot \mathbf{e}^{-\mathbf{x}^{\mathbf{2}}}=\mathbf{x}^{\mathbf{3}}+5$ or $\displaystyle \mathbf{y}=\left(\mathbf{x}^{\mathbf{3}}+5\right) \mathbf{e}^{\mathbf{x}^{\mathbf{2}}}$

Or (b) $\displaystyle \mathrm{x}^{2} d \mathrm{y}+\mathrm{y}(\mathrm{x}+\mathrm{y}) d \mathrm{x}=0 \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=-\frac{\mathrm{y}}{\mathrm{x}}-\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}$

\[\text { Putting } \frac{\mathrm{y}}{\mathrm{x}}=v \Rightarrow \mathrm{y}=v \mathrm{x}, \frac{d \mathrm{y}}{d \mathrm{x}}=v+\mathrm{x} \frac{d v}{d \mathrm{x}} \]

\[v+\mathrm{x} \frac{d v}{d \mathrm{x}}=-v-v^{2} \]

separating the variable and integrating

$$\int \frac{1}{v^{$\displaystyle 2$}+$\displaystyle 2$ v} d v=-\int \frac{1}{\mathrm{x}} d \mathrm{x}

$$$$ \Rightarrow \int \frac{1}{(v+$\displaystyle 1$)^{$\displaystyle 2$}-$\displaystyle 1$} d v=-\int \frac{1}{\mathrm{x}} d \mathrm{x}

$$$\Rightarrow \frac{1}{2} \log \left|\frac{v}{v+2}\right|=\log \left|\frac{C}{\mathrm{x}}\right|$