Given differential equation is

$$\begin{array}{l} \Rightarrow & y^2 d x+\left(x^2-x y+y^2\right) d y =0 \\ \Rightarrow & y^2 d x =-\left(x^2-x y+y^2\right) d y \\ \Rightarrow & y^2 \frac{d x}{d y} =-\left(x^2-x y+y^2\right) \\ \Rightarrow & \frac{d x}{d y} =-\left(\frac{x^2}{y^2}-\frac{x}{y}+1\right)\quad\text{.... (i)} \end{array}$$

which is a homogeneous differential equation.

Put $$\frac{x}{y}=v \text { or } x=v y$$

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

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

$$v+y \frac{d v}{d y}=-\left[v^2-v+1\right]$$

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

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

Given differential equation is

$$\begin{aligned} (x+y)(d x-d y) & =d x+d y \\ \Rightarrow\quad (x+y)\left(1-\frac{d y}{d x}\right) & =1+\frac{d y}{d x} \quad\text{.... (i)}\\ \text{Put}\quad x+y & =z \\ \Rightarrow\quad 1+\frac{d y}{d x} & =\frac{d z}{d x} \end{aligned}$$

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

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

$$\begin{array}{lr} \Rightarrow & 2 z-(z+1) \frac{d z}{d x}=0 \\ \Rightarrow & \frac{d z}{d x}=\frac{2 z}{z+1} \\ \Rightarrow & \left(\frac{z+1}{z}\right) d z=2 d x \end{array}$$

$$\begin{aligned} &\text { On integrating both sides, we get }\\ &\int\left(1+\frac{1}{z}\right) d z=2 \int d x \end{aligned}$$

$$\begin{array}{lrl} \Rightarrow & z+\log z & =2 x-\log C \\ \Rightarrow & (x+y)+\log (x+y) & =2 x-\log C \quad[\because z=x+y]\\ \Rightarrow & 2 x-x-y & =\log C+\log (x+y) \\ \Rightarrow & x-y & =\log |C(x+y)| \\ \Rightarrow & e^{x-y} & =C(x+y) \\ \Rightarrow & (x+y) & =\frac{1}{C} e^{x-y} \\ \Rightarrow & x+y & =K e^{x-y}\quad\left[\because K=\frac{1}{C}\right] \end{array}$$

$$\begin{aligned} \text{Given that,}\quad 2(y+3)-x y \frac{d y}{d x} & =0 \\ \Rightarrow\quad 2(y+3) & =x y \frac{d y}{d x} \\ \Rightarrow\quad 2 \frac{d x}{x} & =\left(\frac{y}{y+3}\right) d y \\ \Rightarrow\quad 2 \cdot \frac{d x}{x} & =\left(\frac{y+3-3}{y+3}\right) d y \\ \Rightarrow\quad 2 \cdot \frac{d x}{x} & =\left(1-\frac{3}{y+3}\right) d y \end{aligned}$$

On integrating both sides, we get

$$2 \log x=y-3 \log (y+3)+C\quad\text{.... (i)}$$

When $x=1$ and $y=-2$, then

$$\begin{aligned} 2 \log 1 & =-2-3 \log (-2+3)+C \\ \Rightarrow\quad 2 \cdot 0 & =-2-3 \cdot 0+C \\ \Rightarrow \quad C & =2 \end{aligned}$$

On substituting the value of $C$ in Eq. (i), we get

$$2 \log x=y-3 \log (y+3)+2$$

$$ \begin{array}{rr} \Rightarrow & 2 \log x+3 \log (y+3)=y+2 \\ \Rightarrow & \log x^2+\log (y+3)^3=(y+2) \\ \Rightarrow & \log x^2(y+3)^3=y+2 \\ \Rightarrow & x^2(y+3)^3=e^{y+2} \end{array}$$

Given differential equation,

$$\begin{array}{ll} \Rightarrow & \frac{d y}{}=\cos x(2-y \operatorname{cosec} x) d x \\ \Rightarrow & \frac{d y}{d x}=\cos x(2-y \operatorname{cosec} x) \\ \Rightarrow & \frac{d y}{d x}=2 \cos x-y \operatorname{cosec} x \cdot \cos x \\ \Rightarrow & \frac{d y}{d x}+y \cot x=2 \cos x \end{array}$$

which is a linear differential equation.

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

$$\begin{aligned} P & =\cot x, Q=2 \cos x \\ \mathrm{IF} & =e^{\int P d x}=e^{\int \cot x d x}=e^{\log \sin x}=\sin x \end{aligned}$$

The general solution is

$$ \begin{array}{ll} & y \cdot \sin x=\int 2 \cos x \cdot \sin x d x+C \\ \Rightarrow & y \cdot \sin x=\int \sin 2 x d x+C \quad[\because \sin 2 x=2 \sin x \cos x]\\ \Rightarrow & y \cdot \sin x=-\frac{\cos 2 x}{2}+C\quad\text{.... (i)} \end{array}$$

When $x=\frac{\pi}{2}$ and $y=2$, then

$$\begin{aligned} &\begin{aligned} 2 \cdot \sin \frac{\pi}{2} & =-\frac{\cos \left(2 \times \frac{\pi}{2}\right)}{2}+C \\ \Rightarrow\quad 2 \cdot 1 & =+\frac{1}{2}+C \\ \Rightarrow\quad 2-\frac{1}{2} & =C \Rightarrow \frac{4-1}{2}=C \\ \Rightarrow\quad C & =\frac{3}{2} \end{aligned}\\ &\text { On substituting the value of } C \text { in Eq. (i), we get }\\ &y \sin x=-\frac{1}{2} \cos 2 x+\frac{3}{2} \end{aligned}$$

Given differential equation is $\quad A x^2+B y^2=1$

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

$$\begin{aligned} 2 A x+2 B y \frac{d y}{d x} & =0 \\ \Rightarrow\quad 2 B y \frac{d y}{d x} & =-2 A x \\ \Rightarrow\quad B y \frac{d y}{d x} & =-A x \Rightarrow \frac{y}{x} \cdot \frac{d y}{d x}=-\frac{A}{B} \end{aligned}$$

$$\begin{aligned} &\text { Again, differentiating w.r.t. } x \text {, we get }\\ &\begin{aligned} & \frac{y}{x} \cdot \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=0 \\ \Rightarrow \quad & \frac{y}{x} \cdot \frac{d^2 y}{d x^2}+\frac{x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)}{x^2}=0 \\ \Rightarrow \quad & x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)=0 \\ \Rightarrow \quad & x y y^{\prime \prime}+x\left(y^{\prime}\right)^2-y y^{\prime}=0 \end{aligned} \end{aligned}$$