Given differential equation is

$$\frac{d x}{d y}+P_1 x=Q_1$$

The general solution is

$$x \cdot \mathrm{IF}=\int Q(\mathrm{IF}) d y+C \text { i.e., } x \mathrm{e}^{\int P d y}=\int Q\left\{e^{\int P d y}\right\} d y+C$$

Given differential equation is

$$x \frac{d y}{d x}+2 y=x^2 \Rightarrow \frac{d y}{d x}+\frac{2 y}{x}=x$$

This equation of the form $\frac{d y}{d x}+P y=Q$.

$$\therefore\quad \mathrm{IF}=e^{\int \frac{2}{x} d x}=e^{2 \log x}=x^2$$

$$\begin{aligned} &\begin{array}{ll} \text { The general solution is }\\ y x^2=\int x \cdot x^2 d x+C \\ \Rightarrow \quad y x^2=\frac{x^4}{4}+C \\ \Rightarrow \quad y=\frac{x^2}{4}+C x^{-2} \end{array} \end{aligned}$$

Given differential equation is

$$\begin{array}{r} \left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0 \\ \Rightarrow\quad \frac{d y}{d x}+\frac{2 x y}{1+x^2}-\frac{4 x^2}{1+x^2}=0 \end{array}$$

$$\begin{aligned} & \Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{4 x^2}{1+x^2} \\ \therefore\quad & \mathrm{IF}=e^{\int \frac{2 x}{1+x^2} d x} \\ & \text { Put } \quad 1+x^2=t \Rightarrow 2 x d x=d t \\ \therefore\quad & \text { IF }=e^{\int \frac{d t}{t}}=e^{\log t}=e^{\log \left(1+x^2\right)}=1+x^2 \end{aligned}$$

The general solution is

$$\begin{aligned} & y \cdot\left(1+x^2\right)=\int\left(1+x^2\right) \frac{.4 x^2}{\left(1+x^2\right)} d x+C \\ \Rightarrow \quad & \left(1+x^2\right) y=\int 4 x^2 d x+C \\ \Rightarrow \quad & \left(1+x^2\right) y=4 \frac{x^3}{3}+C \\ \Rightarrow \quad & y=\frac{4 x^3}{3\left(1+x^2\right)}+C\left(1+x^2\right)^{-1} \end{aligned}$$

$$\begin{aligned} &\text { Given differential equation is }\\ &\begin{array}{rlrl} \Rightarrow & y d x+(x+x y) d y =0 \\ \Rightarrow & y d x+x(1+y) d y =0 \\ \Rightarrow & \frac{d x}{-x} =\left(\frac{1+y}{y}\right) d y \\ \Rightarrow & \int \frac{1}{x} d x =-\int\left(\frac{1}{y}+1\right) d y \quad\text{[on integrating]}\\ \Rightarrow & \log (x) =-\log (y)-y+\log A \\ \Rightarrow & \log (x)+\log (y)+y =\log A \\ \Rightarrow & \log (x y)+y =\log A \\ \Rightarrow & \log x y+\log e^y =\log A \\ \Rightarrow & x y e^y =A \\ \Rightarrow & x y =A e^{-y} \end{array} \end{aligned}$$

Given differential equation is

$$\begin{aligned} \frac{d y}{d x}+y & =\sin x \\ \text { IF } & =\int e^{1 d x}=e^x \end{aligned}$$

The general solution is

$$y \cdot e^x=\int e^x \sin x d x+C\quad\text{.... (i)}$$

$$\begin{aligned} \text{Let}\quad I & =\int e^x \sin x d x \\ I & =\sin x \mathrm{e}^x-\int \cos x \mathrm{e}^x d x \\ & =\sin x \mathrm{e}^x-\cos x \mathrm{e}^x+\int(-\sin x) \mathrm{e}^x d x \\ 2 I & =\mathrm{e}^x(\sin x-\cos x) \\ I & =\frac{1}{2} e^x(\sin x-\cos x) \end{aligned}$$

From Eq. (i),

$$\begin{aligned} y \cdot e^x & =\frac{x}{2}(\sin x-\cos x)+C \\ \Rightarrow\quad y & =\frac{1}{2}(\sin x-\cos x)+C \cdot e^{-x} \end{aligned}$$