The family of concentric circles with centre $(1,2)$ and radius $a$ is given by

$$\begin{aligned} & (x-1)^2+(y-2)^2 & =a^2 \\ &\Rightarrow \quad x^2+1-2 x+y^2+4-4 y =a^2 \\ &\Rightarrow \quad x^2+y^2-2 x-4 y+5 =a^2\quad\text{.... (i)} \end{aligned}$$

On differentiating Eq. (i) w.r.t. $x$, we get

$$\begin{array}{rlr} & 2 x+2 y \frac{d y}{d x}-2-4 \frac{d y}{d x}=0 \\ \Rightarrow & (2 y-4) \frac{d y}{d x}+2 x-2=0 \\ \Rightarrow & (y-2) \frac{d y}{d x}+(x-1)=0 \end{array}$$

Given differential equation is

$$\begin{array}{cc} & y+\frac{d}{d x}(x y)=x(\sin x+\log x) \\ \Rightarrow & y+x \frac{d y}{d x}+y=x(\sin x+\log x) \\ \Rightarrow & x \frac{d y}{d x}+2 y=x(\sin x+\log x) \\ \Rightarrow & \frac{d y}{d x}+\frac{2}{x} y=\sin x+\log x \end{array}$$

which is a linear differential equation.

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

$$\begin{aligned} & P=\frac{2}{x}, Q=\sin x+\log x \\ & \text { IF }=e^{\int \frac{2}{x} \frac{2 x}{x}}=e^{2 \log x}=x^2 \end{aligned}$$

The general solution is

$$ \begin{array}{ll} & y \cdot x^2=\int(\sin x+\log x) x^2 d x+C \\ \Rightarrow & y \cdot x^2=\int\left(x^2 \sin x+x^2 \log x\right) d x+C \\ \Rightarrow & y \cdot x^2=\int x^2 \sin x d x+\int x^2 \log x d x+C \\ \Rightarrow\quad & y \cdot x^2=I_1+I_2+C\quad\text{.... (i)} \end{array}$$

$$ \begin{aligned} &\begin{aligned} \text { Now, } \quad & I_1 =\int x^2 \sin x d x \\ & =x^2(-\cos x)+\int 2 x \cos x d x \\ & =-x^2 \cos x+\left[2 x(\sin x)-\int 2 \sin x d x\right] \\ & I_1 =-x^2 \cos x+2 x \sin x+2 \cos x \quad\text{.... (ii)}\\ \text { and }\quad & I_2 =\int x^2 \log x d x \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & =\log x \cdot \frac{x^3}{3}-\int \frac{1}{x} \cdot \frac{x^3}{3} d x \\ & =\log x \cdot \frac{x^3}{3}-\frac{1}{3} \int x^2 d x \\ & =\log x \cdot \frac{x^3}{3}-\frac{1}{3} \cdot \frac{x^3}{3}\quad\text{.... (iii)} \end{aligned}\\ &\text { On substituting the value of } I_1 \text { and } I_2 \text { in Eq. (i), we get }\\ &\begin{aligned} y \cdot x^2 & =-x^2 \cos x+2 x \sin x+2 \cos x+\frac{x^3}{3} \log x-\frac{1}{9} x^3+C \\ \therefore \quad y & =-\cos x+\frac{2 \sin x}{x}+\frac{2 \cos x}{x^2}+\frac{x}{3} \log x-\frac{x}{9}+C x^{-2} \end{aligned} \end{aligned}$$

Given differential equation is $(1+\tan y)(d x-d y)+2 x d y=0$

on dividing throughout by $d y$, we get

$$\begin{aligned} & (1+\tan y)\left(\frac{d x}{d y}-1\right)+2 x =0 \\ \Rightarrow \quad& (1+\tan y) \frac{d x}{d y}-(1+\tan y)+2 x =0 \\ \Rightarrow \quad& (1+\tan y) \frac{d x}{d y}+2 x =(1+\tan y) \\ \Rightarrow \quad& \frac{d x}{d y}+\frac{2 x}{1+\tan y} =1 \end{aligned}$$

which is a linear differential equation.

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

$$\begin{aligned} P & =\frac{2}{1+\tan y}, Q=1 \\ \text { IF } & =e^{\int \frac{2}{1+\tan y} d y}=e^{\int \frac{2 \cos y}{\cos y+\sin y} d y} \\ & =e^{\int \frac{\cos y+\sin y+\cos y-\sin y}{\cos y+\sin y} d y} \\ & =e^{\int\left(1+\frac{\cos y-\sin y}{\cos y+\sin y}\right) d y}=e^{y+\log (\cos y+\sin y)} \\ & =e^y \cdot(\cos y+\sin y)\quad \left[\because e^{\log x}=x\right] \end{aligned}$$

The general solution is

$$\begin{array}{ll} & \\ & x \cdot e^y(\cos y+\sin y)=\int 1 \cdot e^y(\cos y+\sin y) d y+C \\ \Rightarrow & x \cdot e^y(\cos y+\sin y)=\int e^y(\sin y+\cos y) d y+C \\ \Rightarrow & x \cdot e^y(\cos y+\sin y)=e^y \sin y+C \quad\left[\because \int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)\right] \\ \Rightarrow & x(\sin y+\cos y)=\sin y+C e^{-y} \end{array}$$

Given, $$\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\quad\text{.... (i)}$$

Put $\quad x+y=z$

$$\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d z}{d x}$$

On substituting these values in Eq. (i), we get

$$\begin{array}{ll} & \left(\frac{d z}{d x}-1\right)=\cos z+\sin z \\ \Rightarrow & \frac{d z}{d x}=(\cos z+\sin z+1) \\ \Rightarrow \quad & \frac{d z}{\cos z+\sin z+1}=d x \end{array}$$

$$\begin{aligned} &\text { On integrating both sides, we get }\\ &\begin{aligned} & \int \frac{d z}{\cos z+\sin z+1}=\int 1 d x \\ & \Rightarrow \quad \int \frac{d z}{\frac{1-\tan ^2 z / 2}{1+\tan ^2 z / 2}+\frac{2 \tan z / 2}{1+\tan ^2 z / 2}+1}=\int d x \\ & \Rightarrow \quad \int \frac{d z}{\frac{1-\tan ^2 z / 2+2 \tan z / 2+1+\tan ^2 z / 2}{\left(1+\tan ^2 z / 2\right)}}=\int d x \\ & \Rightarrow \quad \int \frac{\left(1+\tan ^2 z / 2\right) d z}{2+2 \tan ^2 z / 2}=\int d x \\ & \Rightarrow \quad \int \frac{\sec ^2 z / 2 d z}{2(1+\tan z / 2)}=\int d x \\ & \text { Put } 1+\tan z / 2=t \Rightarrow \quad\left(\frac{1}{2} \sec ^2 z / 2\right) d z=d t \\ & \Rightarrow \quad \int \frac{d t}{t}=\int d x \\ & \Rightarrow \quad \log |t|=x+C \\ & \Rightarrow \quad \log |1+\tan z / 2|=x+C \\ & \Rightarrow \quad \log \left|1+\tan \frac{(x+y)}{2}\right|=x+C \end{aligned} \end{aligned}$$

Given, $$\frac{d y}{d x}-3 y=\sin 2 x$$

which is a linear differential equation.

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

$$\begin{aligned} P & =-3, Q=\sin 2 x \\ I F & =e^{-3 \int d x}=e^{-3 x} \end{aligned}$$

The general solution is

$$y\,.\,{e^{ - 3x}} = \int {\mathop {\sin 2x}\limits_I \mathop {{e^{ - 3x}}}\limits_{II} dx} $$

Let $$y\,.\,{e^{ - 3x}} = I\quad\text{.... (i)}$$

$$\therefore\quad I = \int {\mathop {{e^{ - 3x}}}\limits_{II} \mathop {\sin 2x}\limits_I } $$

$$\begin{array}{ll} \Rightarrow & I=\sin 2 x\left(\frac{e^{-3 x}}{-3}\right)-\int 2 \cos 2 x\left(\frac{e^{-3 x}}{-3}\right) d x+C_1 \\ \Rightarrow & I=-\frac{1}{3} e^{-3 x} \sin 2 x+\frac{2}{3} \int e^{-3 x} \cos 2 x d x+C_1 \\ \Rightarrow & I=-\frac{1}{3} e^{-3 x} \sin 2 x+\frac{2}{3}\left(\cos 2 x \frac{e^{-3 x}}{-3}-\int(-2 \sin 2 x) \frac{e^{-3 x}}{-3} d x\right)+C_1+C_2 \\ \Rightarrow & I=-\frac{1}{3} e^{-3 x} \sin 2 x-\frac{2}{9} \cos 2 x e^{-3 x}-\frac{4}{9} I+C^{\prime} \quad \quad \quad\left[\text { where }, C^{\prime}=C_1+C_2\right] \\ \Rightarrow & I+\frac{41}{9} 2=+e^{-3 x}\left(-\frac{1}{3} \sin 2 x-\frac{2}{9} \cos 2 x\right)+C^{\prime} \end{array}$$

$$\begin{aligned} & \Rightarrow \quad \frac{13}{9} I=e^{-3 x}\left(-\frac{1}{3} \sin 2 x-\frac{2}{9} \cos 2 x\right)+C^{\prime} \\ & \Rightarrow \quad I=\frac{9}{13} \mathrm{e}^{-3 x}\left(-\frac{1}{3} \sin 2 x-\frac{2}{9} \cos 2 x\right)+C \quad \left[\text { where } C=\frac{9 C^{\prime}}{13}\right]\\ & \Rightarrow \quad I=\frac{3}{13} e^{-3 x}\left(-\sin 2 x-\frac{2}{3} \cos 2 x\right)+C \\ & \Rightarrow \quad=\frac{3}{13} e^{-3 x} \frac{(-3 \sin 2 x-2 \cos 2 x)}{3}+C \\ & \Rightarrow \quad=\frac{e^{-3 x}}{13}(-3 \sin 2 x-2 \cos 2 x)+C \\ & \Rightarrow \quad I=-\frac{e^{-3 x}}{13}(2 \cos 2 x+3 \sin 2 x)+C \end{aligned}$$

$$\begin{aligned} &\text { On substituting the value of } I \text { in Eq. (i), we get }\\ &\begin{aligned} y \cdot e^{-3 x} & =-\frac{e^{-3 x}}{13}(2 \cos 2 x+3 \sin 2 x)+C \\ \Rightarrow \quad y & =-\frac{1}{13}(2 \cos 2 x+3 \sin 2 x)+C e^{3 x} \end{aligned} \end{aligned}$$