$$\begin{aligned} &\text { It is given that, the slope of tangent to the curve at point }(x, y) \text { is } \frac{x^2+y^2}{2 x y} \text {. }\\ &\begin{array}{lrl} \therefore & \left(\frac{d y}{d x}\right)_{(x, y)} & =\frac{x^2+y^2}{2 x y} \\ \Rightarrow & \frac{d y}{d x} & =\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right) \end{array} \end{aligned}$$

which is homogeneous differential equation.

$$\begin{array}{ll} & \begin{aligned} \text { Put } y & =v x \\ \Rightarrow \quad & \frac{d y}{d x} \end{aligned}=v+x \frac{d v}{d x} \end{array}$$

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

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

On integrating both sides, we get

$$\int \frac{2 v}{1-v^2} d v=\int \frac{d x}{x}$$

Put $1-v^2=t$ in LHS, we get

$$-2 v d v=d t$$

$$\begin{array}{lr} \Rightarrow & -\int \frac{d t}{t}=\int \frac{d x}{x} \\ \Rightarrow & -\log t=\log x+\log C \\ \Rightarrow & -\log \left(1-v^2\right)=\log x+\log C \\ \Rightarrow & -\log \left(1-\frac{y^2}{x^2}\right)=\log x+\log C \\ \Rightarrow & \log \left(\frac{x^2-y^2}{x^2}\right)=\log x+\log C \\ \Rightarrow & \left.\frac{x^2}{x^2-y^2}\right)=\log x+\log C \\ \Rightarrow & \frac{x^2}{x^2-y^2}=C x\quad\text{.... (i)} \end{array}$$

Since, the curve passes through the point $(2,1)$.

$$\therefore \quad \frac{(2)^2}{(2)^2-(1)^2}=C(2) \Rightarrow C=\frac{2}{3}$$

So, the required solution is $2\left(x^2-y^2\right)=3 x$.

It is given that, slope of tangent to the curve at any point $(x, y)$ is $\frac{y-1}{x^2+x}$.

$$\begin{array}{lr} \therefore & \left(\frac{d y}{d x}\right)_{(x, y)}=\frac{y-1}{x^2+x} \\ \Rightarrow & \frac{d y}{d x}=\frac{y-1}{x^2+x} \\ \Rightarrow & \frac{d y}{y-1}=\frac{d x}{x^2+x} \end{array}$$

On integrating both sides, we get

$$\begin{array}{ll} \Rightarrow & \int \frac{d y}{y-1}=\int \frac{d x}{x^2+x} \\ \Rightarrow & \int \frac{d y}{y-1}=\int \frac{d x}{x(x+1)} \\ \Rightarrow & \int \frac{d y}{y-1}=\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x \\ \Rightarrow & \log (y-1)=\log x-\log (x+1)+\log C \\ \Rightarrow & \log (y-1)=\log \left(\frac{x C}{x+1}\right) \end{array}$$

$$\begin{aligned} &\text { Since, the given curve passes through point }(1,0) \text {. }\\ &\therefore \quad 0-1=\frac{1 \cdot C}{1+1} \Rightarrow C=-2 \end{aligned}$$

$$\begin{aligned} & \text { The particular solution is } \quad y-1=\frac{-2 x}{x+1} \\ &\begin{aligned} \Rightarrow \quad (y-1)(x+1) & =-2 x \\ \Rightarrow \quad(y-1)(x+1)+2 x & =0 \end{aligned} \end{aligned}$$

Slope of tangent to the curve $=\frac{d y}{d x}$

and difference of abscissa and ordinate $=x-y$

According to the question,

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

$$\begin{aligned} & \text { Put } \quad x-y=z \\ & \Rightarrow\quad 1-\frac{d y}{d x}=\frac{d z}{d x} \\ & \Rightarrow \quad\frac{d y}{d x}=1-\frac{d z}{d x} \end{aligned}$$

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

$$\begin{array}{rlrl} & 1-\frac{d z}{d x} =z^2 \\ \Rightarrow & 1-z^2 =\frac{d z}{d x} \\ \Rightarrow & d x =\frac{d z}{1-z^2} \end{array}$$

On integrating both sides, we get

$$ \begin{aligned} \int d x & =\int \frac{d z}{1-z^2} \\ \Rightarrow\quad x & =\frac{1}{2} \log \left|\frac{1+z}{1-z}\right|+C \\ \Rightarrow\quad \mathrm{t} x & =\frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right|+C\quad\text{... (ii)} \end{aligned}$$

Since, the curve passes through the origin.

$$\begin{array}{lc} \therefore & 0=\frac{1}{2} \log \left|\frac{1+0-0}{1-0+0}\right|+C \\ \Rightarrow & C=0 \end{array}$$

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

$$\begin{array}{rlrl} & x =\frac{1}{2} \log \left|\frac{1+x-y}{1-x+y}\right| \\ \Rightarrow & 2 x =\log \left|\frac{1+x-y}{1-x+y}\right| \\ \Rightarrow \quad & e^{2 x} =\left|\frac{1+x-y}{1-x+y}\right| \\ \Rightarrow \quad & (1-x+y) e^{2 x} =1+x-y \end{array}$$

The below figure obtained by the given information

Let the coordinate of the point $P$ is $(x, y)$. It is given that, $P$ is mid-point of $A B$.

So, the coordinates of points $A$ and $B$ are $(2 x, 0)$ and $(0,2 y)$, respectively.

$\therefore \quad$ Slope of $A B=\frac{0-2 y}{2 x-0}=-\frac{y}{x}$

Since, the segment $A B$ is a tangent to the curve at $P$.

$$\begin{array}{ll} \therefore & \frac{d y}{d x}=-\frac{y}{x} \\ \Rightarrow & \frac{d y}{y}=-\frac{d x}{x} \end{array}$$

On integrating both sides, we get

$$\begin{aligned} & \log y=-\log x+\log C \\ & \log y=\log \frac{C}{x}\quad\text{.... (i)} \end{aligned}$$

Since, the given curve passes through $(1,1)$.

$$\begin{array}{ll} \therefore & \log 1=\log \frac{C}{1} \\ \Rightarrow & 0=\log C \\ \Rightarrow & C=1 \end{array}$$

$$\begin{array}{lc} \therefore & \log y=\log \frac{1}{x} \\ \Rightarrow & y=\frac{1}{x} \\ \Rightarrow & x y=1 \end{array}$$

$$\begin{array}{ll} \text { Given, } & x \frac{d y}{d x}=y(\log y-\log x+1) \\ \Rightarrow & x \frac{d y}{d x}=y \log \left(\frac{y}{x}+1\right) \\ \Rightarrow & \frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)\quad\text{.... (i)} \end{array}$$

which is a homogeneous equation.

$$\begin{aligned} \text { Put } & \frac{y}{x} =v \text { or } y=v x \\ \therefore & \frac{d y}{d x} =v+x \frac{d v}{d x} \end{aligned}$$

$$\begin{aligned} &\text { On substituting these values in Eq.(i), we get }\\ &\begin{array}{rlrl} \Rightarrow & v+x \frac{d v}{d x} =v(\log v+1) \\ \Rightarrow & x \frac{d v}{d x} =v(\log v+1- \\ \Rightarrow & x \frac{d v}{d x} =v(\log v) \\ \Rightarrow & \frac{d v}{v \log v}=\frac{d x}{x} \end{array} \end{aligned}$$

On integrating both sides, we get

$$\int \frac{d v}{v \log v}=\int \frac{d x}{x}$$

On putting $\log v=u$ in LHS integral, we get

$$\begin{aligned} & \frac{1}{v} \cdot d v=d u \\ & \int \frac{d u}{u}=\int \frac{d x}{x} \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad & \log u =\log x+\log C \\ \Rightarrow \quad & \log u =\log C x \\ \Rightarrow \quad & u =C x \\ \Rightarrow \quad & \log v =C x \\ \Rightarrow \quad & \log \left(\frac{y}{x}\right) =C x \end{aligned}$$