$$\begin{array}{r} \text{Given that,}\quad \frac{d y}{d x} & =2^{y-x} \\ \Rightarrow\quad\frac{d y}{d x} & =\frac{2^y}{2^x} \\ \Rightarrow\quad\frac{d y}{2^y} & =\frac{d x}{2^x} \end{array} \quad\left[\because a^{m-n}=\frac{a^m}{a^n}\right]$$

$$\begin{aligned} &\text { On integrationg both sides, we get }\\ &\begin{array}{l} & \int 2^{-y} d y =\int 2^{-x} d x \\ \Rightarrow & \frac{-2^{-y}}{\log 2} =\frac{-2^{-x}}{\log 2}+C \\ \Rightarrow & -2^{-y}+2^{-x}=+C \log 2 \\ \Rightarrow & 2^{-x}-2^{-y}=-C \log 2 \\ \Rightarrow & 2^{-x}-2^{-y}=K\quad\text { [where, } K=+C \log 2 \text { ] } \end{array} \end{aligned}$$

Since, the family of all non-vertical line is $y=m x+c$, where $m \neq \tan \frac{\pi}{2}$.

On differentiating w.r.t. $x$, we get

$$\frac{d y}{d x}=m$$

Again, differentiating w.r.t. $x$, we get

$$\frac{d^2 y}{d x^2}=0$$

$$\begin{aligned} & \text { Given that, } \quad \frac{d y}{d x}=e^{-2 y} \Rightarrow \quad \frac{d y}{e^{-2 y}}=d x \\ & \Rightarrow \quad \int e^{2 y} d y=\int d x \Rightarrow \frac{e^{2 y}}{2}=x+C\quad\text{..... (i)} \end{aligned}$$

$$\begin{aligned} &\text { When } x=5 \text { and } y=0 \text {, then substituting these values in Eq. (i), we get }\\ &\begin{array}{l} & \frac{e^0}{2}=5+C \\ \Rightarrow & \frac{1}{2} =5+C \quad \Rightarrow \quad C=\frac{1}{2}-5=-\frac{9}{2} \\ \text { Eq. (i) becomes } & e^{2 y} =2 x-9 \\ \text { When } y=3 \text {, then } & e^6 =2 x-9 \Rightarrow 2 x=e^6+9 \\ \therefore & x =\frac{\left(e^6+9\right)}{2} \end{array} \end{aligned}$$

Given differential equation is

$$\left(x^2-1\right) \frac{d y}{d x}+2 x y=\frac{1}{x^2-1}$$

$\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{x^2-1}\right) y=\frac{1}{\left(x^2-1\right)^2}$

which is a linear differential equation.

On comparing it with $\frac{d y}{d x}+P y=Q$, we get

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

The complete solution is

$$\begin{aligned} y \cdot \mathrm{IF} & =\int Q \cdot \mathrm{IF}+K \\ \Rightarrow\quad y \cdot\left(x^2-1\right) & =\int \frac{1}{\left(x^2-1\right)^2} \cdot\left(x^2-1\right) d x+K \\ \Rightarrow\quad y \cdot\left(x^2-1\right) & =\int \frac{d x}{\left(x^2-1\right)}+K \\ \Rightarrow\quad y \cdot\left(x^2-1\right) & =\frac{1}{2} \log \left(\frac{x-1}{x+1}\right)+K \end{aligned}$$

Given that, $$\frac{d y}{d x}+2 x y=y$$

$$\begin{aligned} &\begin{array}{ll} \Rightarrow & \frac{d y}{d x}+2 x y-y=0 \\ \Rightarrow & \frac{d y}{d x}+(2 x-1) y=0 \end{array}\\ &\text { which is a linear differential equation. } \end{aligned}$$

$$\begin{aligned} &\text { On comparing it with } \frac{d y}{d x}+P y=Q \text {, we get }\\ &\begin{aligned} P & =(2 x-1), Q=0 \\ \mathrm{IF} & =e^{\int P d x}=e^{\int(2 x-1) d x} \\ & =e^{\left(\frac{2 x^2}{2}-x\right)}=e^{x^2-x} \end{aligned} \end{aligned}$$

The complete solution is

$$\begin{array}{l} & y \cdot e^{x^2-x} =\int Q \cdot e^{x^2-x} d x+C \\ \Rightarrow & y \cdot e^{x^2-x} =0+C \\ \Rightarrow & y =C e^{x-x^2} \end{array}$$