The given differential equation is $$y = x\frac{{dy}}{{dx}} + \frac{{dx}}{{dy}}$$
Multiplying both the sides by $$\frac{{dy}}{{dx}}$$  we get
$$\eqalign{ & \left( {\frac{{dy}}{{dx}}} \right)y = x{\left( {\frac{{dy}}{{dx}}} \right)^2} + 1 \cr & \Rightarrow x{\left( {\frac{{dy}}{{dx}}} \right)^2} - y\left( {\frac{{dy}}{{dx}}} \right) + 1 = 0 \cr} $$
Hence, order and degree of differential equation are $$1$$ and $$2.$$

Let $$P\left( {x,\,y} \right)$$  be any point on the curve, $$PM$$  the perpendicular to $$x$$-axis $$PT$$  the tangent at $$P$$ meeting the axis of $$x$$ at $$T.$$ As given $$OT = 2\,OM =2x.$$    Equation of the tangent at $$P\left( {x,\,y} \right)$$  is $$Y - y = \frac{{dy}}{{dx}}\left( {X - x} \right)$$
It intersects the axis of $$x$$ where $$Y = 0$$
i.e., $$ - y = \frac{{dy}}{{dx}}\left( {X - x} \right){\text{ or }}X = x - y\frac{{dy}}{{dx}} = OT$$
Hence, $$x - y\frac{{dy}}{{dx}} = 2x{\text{ or }}\frac{{dx}}{x} + \frac{{dy}}{y} = 0$$
Integrating, $$\log \,x + \log \,y = \log \,C{\text{ i}}{\text{.e}}{\text{., }}xy = C$$
This passes through $$\left( {1,\,2} \right),$$
$$\therefore \,C = 2$$
Hence the required curve is $$xy = 2$$

$$\eqalign{ & {\text{Circle }}{\left( {x - h} \right)^2} + {\left( {y - 2} \right)^2} = 25 \cr & \Rightarrow 2\left( {x - h} \right) + 2\left( {y - 2} \right)y' = 0 \cr & \Rightarrow 2\left( {y - 2} \right)y' = - 2\left( {x - h} \right) \cr & \Rightarrow {\left( {y - 2} \right)^2}y{'^2} = {\left( {x - h} \right)^2} \cr & \Rightarrow {\left( {y - 2} \right)^2}y{'^2} = 25 - {\left( {y - 2} \right)^2} \cr} $$

Divide the equation by $$y{\left( {\log \,y} \right)^2}$$
$$\eqalign{ & \frac{1}{{y{{\left( {\log \,y} \right)}^2}}}\frac{{dy}}{{dx}} + \frac{1}{{\log \,y}}.\frac{1}{x} = \frac{1}{{{x^2}}} \cr & {\text{Put }}\frac{1}{{\log \,y}} = z \cr & \Rightarrow \frac{{ - 1}}{{y{{\left( {\log \,y} \right)}^2}}}\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}} \cr & {\text{Thus, we get, }} - \frac{{dz}}{{dx}} + \frac{1}{x}.z = \frac{1}{{{x^2}}},{\text{ linear in }}z \cr & \Rightarrow \frac{{dz}}{{dx}} + \left( { - \frac{1}{x}} \right)z = - \frac{1}{{{x^2}}}\,; \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{ - \int {\frac{1}{x}dx} }} = {e^{ - \log \,x}} = \frac{1}{x} \cr & \therefore \,{\text{The solution is,}} \cr & z\left( {\frac{1}{x}} \right) = \int {\frac{{ - 1}}{{{x^2}}}} \left( {\frac{1}{x}} \right)dx + c \cr & \Rightarrow \frac{1}{{\log \,y}}\left( {\frac{1}{x}} \right) = \frac{{ - {x^{ - 2}}}}{{ - 2}} + c \cr & \Rightarrow x = \log \,y\left( {c{x^2} + \frac{1}{2}} \right) \cr} $$

$$\eqalign{ & {\text{I}}{\text{.F}}{\text{.}} = {e^{ - x}} \cr & \therefore \,y{e^{ - x}} = \int {{e^{ - x}}} \left( {\cos \,x - \sin \,x} \right)dx \cr & {\text{Put }} - x = t \cr} $$

$$\eqalign{ & = - \int {{e^t}\left( {\cos \,t + \sin \,t} \right)dt} \cr & = - {e^t}\sin \,t + c \cr & y{e^{ - x}} = {e^{ - x}}\sin \,x + c \cr} $$
Since, $$y$$ is bounded when $$x \to \infty \Rightarrow c = 0$$
$$\eqalign{ & \therefore \,y = \sin \,x \cr & {\text{Area}} = \int_0^{\frac{\pi }{4}} {\left( {\cos \,x - \sin \,x} \right)dx = \sqrt 2 - 1} \cr} $$

$$\eqalign{ & {x^2} = {e^{{{\left( {\frac{x}{y}} \right)}^{ - 1}}\left( {\frac{{dy}}{{dx}}} \right)}} \Rightarrow {x^2} = {e^{\left( {\frac{y}{x}} \right)\left( {\frac{{dy}}{{dx}}} \right)}} \cr & \Rightarrow \ln \,{x^2} = \frac{y}{x}\frac{{dy}}{{dx}}{\text{ or }}\int {x\,\ln \,{x^2}dx} = \int {y\,dy} \cr & {\text{Put }}{x^2} = t\, \Rightarrow 2x\,dx = dt \cr & \therefore \frac{1}{2}\int {\ln \,t\,dt = \frac{{{y^2}}}{2}} \cr & C + t\,\ln \,t - t = {y^2} \cr & {\text{or }}{y^2} = {x^2}\left( {\ln {x^2} - 1} \right) + C \cr} $$

$$\eqalign{ & ydx + \left( {x + {x^2}y} \right)dy = 0 \cr & \Rightarrow \frac{{dx}}{{dy}} = - \frac{x}{y} - {x^2}\,\,\,\, \Rightarrow \frac{{dx}}{{dy}} + \frac{x}{y} = - {x^2} \cr} $$
It is Bernoullis form. Divide by $${x^2}$$
$$\eqalign{ & {x^{ - 2}}\frac{{dx}}{{dy}} + {x^{ - \,1}}\left( {\frac{1}{y}} \right) = - 1 \cr & {\text{Put }}{x^{ - \,1}} = t,\,\, - {x^{ - \,2}}\frac{{dx}}{{dy}} = \frac{{dt}}{{dy}} \cr & {\text{We get,}} \cr & - \frac{{dt}}{{dy}} + t\left( {\frac{1}{y}} \right) = - 1\,\,\,\,\,\, \Rightarrow \frac{{dt}}{{dy}} - \left( {\frac{1}{y}} \right)t = 1 \cr} $$
It is linear in $$t.$$
Integrating factor $$ = {e^{\int { - \,\frac{1}{y}dy} }} = {e^{ - \,\log \,y}} = {y^{ - 1}}$$
$$\therefore $$ Solution is $$t\left( {{y^{ - \,1}}} \right) = \int {\left( {{y^{ - \,1}}} \right)} dy + c$$
$$ \Rightarrow \frac{1}{x}.\frac{1}{y} = \log \,y + c \Rightarrow \log \,y - \frac{1}{{xy}} = c$$

Since order of the highest derivative in the given differential equation is $$1$$ and exponent of the derivative is also $$1$$ therefore degree and order is $$\left( {1,\,1} \right).$$

Given that $$y = y\left( x \right)$$
and $$x\,\cos \,y + y\,\cos \,x = \pi \,.....(1)$$
For $$x=0$$   in (1) we get $$y = \pi $$
Differentiating (1) with respect to $$x,$$ we get
$$\eqalign{ & - x\,\sin \,y.y' + \cos \,y + y'\,\cos \,x - y\,\sin \,x = 0 \cr & \Rightarrow y' = \frac{{y\,\sin \,x - \cos \,y}}{{\cos \,x - x\,\sin \,y}}\,.....(2) \cr & \Rightarrow y'\left( 0 \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Using }}y\left( 0 \right) = \pi } \right) \cr} $$
Differentiating (2) with respect to $$x,$$ we get
$$\eqalign{ & y'' = \frac{{\left( {y'\sin \,x + y\,\cos \,x + \sin \,y.y'} \right)\left( {\cos \,x - x\,\sin \,y} \right) - \left( { - \sin \,x - \sin \,y - x\,\cos \,yy'} \right)\left( {y\,\sin \,x - \cos \,y} \right)}}{{{{\left( {\cos \,x - x\,\sin \,y} \right)}^2}}} \cr & \Rightarrow y''\left( 0 \right) = \frac{{\pi \left( 1 \right) - 1}}{1} = \pi - 1 \cr} $$

Given differential equation is
$$\eqalign{ & a\left( {x\frac{{dy}}{{dx}} + 2y} \right) = xy\frac{{dy}}{{dx}} \cr & \Rightarrow ax\frac{{dy}}{{dx}} - xy\frac{{dy}}{{dx}} = - 2ay \cr & \Rightarrow \left( {xy - ax} \right)\frac{{dy}}{{dx}} = 2ay \cr & \Rightarrow x\left( {y - a} \right)\frac{{dy}}{{dx}} = 2ay \cr & \Rightarrow x\left( {y - a} \right)dy = 2ay\,dx \cr & \Rightarrow \frac{{\left( {y - a} \right)}}{y}dy = \frac{{2a}}{x}dx \cr & \Rightarrow \left( {1 - \frac{a}{y}} \right)dy = \frac{{2a}}{x}dx \cr & dy - \frac{a}{y}dy = \frac{{2a}}{x}dx \cr & {\text{Integrate on both side}} \cr & \int {dy - a} \int {\frac{1}{y}dy} = 2a\int {\frac{1}{x}dx} \cr & y - a\,\log \,y = 2a\,\log \,x + \log \,c \cr & \Rightarrow y = a\,\log \,{x^2}yc \Rightarrow {x^2}y = k{e^{\frac{y}{a}}} \cr} $$