Given that, $$y=\left(\sin ^{-1} x\right)^2+A \cos ^{-1} x+B$$

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

$$\begin{aligned} \frac{d y}{d x} & =\frac{2 \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{(-A)}{\sqrt{1-x^2}} \\ \Rightarrow \quad \sqrt{1-x^2} \frac{d y}{d x} & =2 \sin ^{-1} x-A \end{aligned}$$

$$\begin{aligned} &\text { Again, differentiating w.r.t. } x \text {, we get }\\ &\sqrt{1-x^2} \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot \frac{-2 x}{2 \sqrt{1+x^2}}=\frac{2}{\sqrt{1-x^2}} \end{aligned}$$

$$\begin{array}{lr} \Rightarrow & \left(1-x^2\right) \frac{d^2 y}{d x^2}-\frac{x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} \frac{d y}{d x}=2 \\ \Rightarrow & \left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=2 \\ \Rightarrow & \left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-2=0 \end{array}$$

which is the required differential equation.

It is given that, circles pass through origin and their centreslie on $Y$-axis. Let $(0, k)$ be the centre of the circle and radius is $k$.

So, the equation of circle is

$$\begin{aligned} (x-0)^2+(y-k)^2 & =k^2 \\ \Rightarrow\quad x^2+(y-k)^2 & =k^2 \\ \Rightarrow\quad x^2+y^2-2 k y & =0 \\ \Rightarrow\quad \frac{x^2+y^2}{2 y} & =k\quad\text{.... (i)} \end{aligned}$$

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

$\frac{2 y\left(2 x+2 y \frac{d y}{d x}\right)-\left(x^2+y^2\right) \frac{2 d y}{d x}}{4 y^2}=0$

$$\begin{array}{lr} \Rightarrow & 4 y\left(x+y \frac{d y}{d x}\right)-2\left(x^2+y^2\right) \frac{d y}{d x}=0 \\ \Rightarrow & 4 x y+4 y^2 \frac{d y}{d x}-2\left(x^2+y^2\right) \frac{d y}{d x}=0 \\ \Rightarrow & {\left[4 y^2-2\left(x^2+y^2\right)\right] \frac{d y}{d x}+4 x y=0} \\ \Rightarrow & \left(4 y^2-2 x^2-2 y^2\right) \frac{d y}{d x}+4 x y=0 \\ \Rightarrow & \left(2 y^2-2 x^2\right) \frac{d y}{d x}+4 x y=0 \\ \Rightarrow & \left(y^2-x^2\right) \frac{d y}{d x}+2 x y=0 \\ \Rightarrow & \left(x^2-y^2\right) \frac{d y}{d x}-2 x y=0 \end{array}$$

Given that, $$\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2$$

$$\Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{1+x^2} \cdot y=\frac{4 x^2}{1+x^2}$$

which is a linear differential equation.

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

$$\begin{aligned} & P=\frac{2 x}{1+x^2}, Q=\frac{4 x^2}{1+x^2} \\ \therefore\quad & I F=e^{\int P d x}=e^{\int \frac{2 x}{1+x^2} d x} \end{aligned}$$

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

The general solution is

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

Since, the curve passes through origin, then substituting

$x=0$ and $y=0$ in Eq. (i), we get $C=0$

The required equation of curve is

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

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

$\Rightarrow \quad \frac{d y}{d x}=1+\frac{y}{x}+\frac{y^2}{x^2}\quad\text{.... (i)}$

Let, $$f(x, y)=1+\frac{y}{x}+\frac{y^2}{x^2}$$

$$\begin{aligned} f(\lambda x, \lambda y) & =1+\frac{\lambda y}{\lambda x}+\frac{\lambda^2 y^2}{\lambda^2 x^2} \\ f(\lambda x, \lambda y) & =\lambda^0\left(1+\frac{y}{x}+\frac{y^2}{x^2}\right) \\ & =\lambda^0 f(x, y) \end{aligned}$$

which is homogeneous expression of degree 0.

Put $$y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$$

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

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

$$\begin{array}{ll} \Rightarrow & x \frac{d v}{d x}=1+v+v^2-v \\ \Rightarrow & x \frac{d v}{d x}=1+v^2 \\ \Rightarrow & \frac{d v}{1+v^2}=\frac{d x}{x} \end{array}$$

$$\begin{aligned} &\text { On integrating both sides, we get }\\ &\begin{aligned} \tan ^{-1} v & =\log |x|+C \\ \Rightarrow\quad \tan ^{-1}\left(\frac{y}{x}\right) & =\log |x|+C \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { Given, differential equation is }\\ &\Rightarrow \quad \begin{aligned} \left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x} & =0 \\ \left(1+y^2\right) & =-\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x} \\ \left(1+y^2\right) \frac{d x}{d y} & =-x+e^{\tan ^{-1} y} \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & \left(1+y^2\right) \frac{d x}{d y}+x=e^{\tan ^{-1} y} \\ \Rightarrow & \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{e^{\tan ^{-1} y}}{1+y^2}\quad \text { [dividing throughout by }\left(1+y^2\right) \text { ] } \end{array}$$

which is a linear differential equation.

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

$$\begin{aligned} & P=\frac{1}{1+y^2}, Q=\frac{e^{\tan ^{-1} y}}{1+y^2} \\ & \text { IF }=e^{\int P d y}=e^{\int \frac{1}{1+y^2} d y}=e^{\tan ^{-1} y} \end{aligned}$$

$$\begin{aligned} & \text { The general solution is } \quad x \cdot e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y}}{1+y^2} \cdot e^{\tan ^{-1} y} d y+C \\ & \Rightarrow \quad x \cdot e^{\tan ^{-1} y}=\int \frac{\left(e^{\tan ^{-1} y}\right)^2}{1+y^2} \cdot d y+C \end{aligned}$$

$$\begin{array}{ll} \text { Put } \tan ^{-1} y=t \Rightarrow & \frac{1}{1+y^2} d y=d t \\ \therefore & x \cdot e^{\tan ^{-1} y}=\int e^{2 t} d t+C \end{array}$$

$$ \begin{array}{ll} \Rightarrow & x \cdot e^{\tan ^{-1} y}=\frac{1}{2} e^{2 \tan ^{-1} y}+C \\ \Rightarrow & 2 x \mathrm{e}^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+2 C \\ \Rightarrow & 2 x \mathrm{e}^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+K\quad[\because K=2 C] \end{array}$$