The equation of the family is $${y^2} = 4a\left( {x - b} \right)$$    where $$a,\,b$$  are arbitrary constants.
$$\therefore 2y\frac{{dy}}{{dx}} = 4a\,\,{\text{or}}\,{\text{ }}{\left( {\frac{{dy}}{{dx}}} \right)^2} + y\frac{{{d^2}y}}{{d{x^2}}} = 0$$
So, degree $$=1$$  and order $$=2$$

Given differential equation is $$\frac{{dy}}{{dx}} + \frac{{2yx}}{{1 + {x^2}}} = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}$$       which is a linear.
Differential equation of the form: $$\frac{{dy}}{{dx}} + Py = Q$$
On comparing, we have
$$\eqalign{ & P = \frac{{2x}}{{1 + {x^2}}}{\text{ and }}Q = \frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{{2x}}{{1 + {x^2}}}dx} }} = {e^{\log \left( {1 + {x^2}} \right)}} = \left( {1 + {x^2}} \right) \cr} $$
$$\therefore $$  Solution is
$$\eqalign{ & y\left( {1 + {x^2}} \right) = \int {\frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}} \left( {1 + {x^2}} \right)dx + c \cr & \Rightarrow y\left( {1 + {x^2}} \right) = \int {\frac{1}{{\left( {1 + {x^2}} \right)}}} dx + c \cr & \Rightarrow y\left( {1 + {x^2}} \right) = {\tan ^{ - 1}}x + c \cr} $$

Given, $$\frac{{dy}}{{dx}} + \left( {\frac{1}{{x\,\log \,x}}} \right)y = 2$$
$$\eqalign{ & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{1}{{x\,\log \,x}}\,dx} }} = {e^{\log \left( {\log \,x} \right)}} = \log \,x \cr & y.\log \,x = \int {2\,\log \,x\,dx + c} \cr & y.\log \,x = 2\left[ {x\,\log \,x - x} \right] + c \cr & {\text{Put }}\,x = 1,\,\,y.0 = - 2 + c\,\, \Rightarrow c = 2 \cr & {\text{Put }}x = e \cr & y\,\log \,e = 2e\left( {\log \,e - 1} \right) + c\,\,\, \Rightarrow y\left( e \right) = c = 2 \cr} $$

$$\eqalign{ & \frac{{xdy}}{{dx}} = y\left( {\log \,y - \log \,x + 1} \right) \cr & \frac{{dy}}{{dx}} = \frac{y}{x}\left( {\log \left( {\frac{y}{x}} \right) + 1} \right) \cr & {\text{Put }}y = vx \cr & \frac{{dy}}{{dx}} = v + \frac{{xdv}}{{dx}} \Rightarrow v + \frac{{xdv}}{{dx}} = v\left( {\log \,v + 1} \right) \cr & \frac{{xdv}}{{dx}} = v\,\log \,v \Rightarrow \frac{{dv}}{{v\,\log \,v}} = \frac{{dx}}{x} \cr & {\text{Put }}\log \,v = z \Rightarrow \frac{1}{v}dv = dz \Rightarrow \frac{{dz}}{z} = \frac{{dx}}{x} \cr & \Rightarrow \ln \,z = \ln \,x + \ln \,c \cr & x = cx\,\,\,\,{\text{or, }}\log \,v = cx\,\,\,{\text{or, }}\log \left( {\frac{y}{x}} \right) = cx \cr} $$

The given differential equation is $$\sin \,x\,\cos \,y\,dx + \cos \,x\,\sin \,y\,dy = 0$$
dividing by $$\cos \,x\,\cos \,y$$
$$ \Rightarrow \frac{{\sin \,x}}{{\cos \,x}}dx + \frac{{\sin \,y}}{{\cos \,y}}dy = 0$$
Integrating,
$$\eqalign{ & \int {\tan \,x\,dx} + \int {\tan \,y\,dy} = \log \,c \cr & {\text{or }}\log \,\sec \,x\,\sec \,y = \log \,c \cr & {\text{or }}\sec \,x\,\sec \,y = c \cr} $$
curve passes through the point $$\left( {0,\,\frac{\pi }{4}} \right)$$
$$\sec \,0\,\sec \frac{\pi }{4} = c = \sqrt 2 $$
Hence, the required equation of the curve is $$\sec \,x\,\sec \,y = \sqrt 2 .$$

$$\eqalign{ & {\text{Put }}xy = v \cr & \therefore \,y + x\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} \Rightarrow \frac{{dv}}{{dx}} = x\frac{{\phi \left( v \right)}}{{\phi '\left( v \right)}} \cr & \therefore \,\frac{{\phi '\left( v \right)}}{{\phi \left( v \right)}}dv = x\,dx \cr & {\text{Integrating, we get}} \cr & \log \,\phi \left( v \right) = \frac{{{x^2}}}{2} + \log \,k \cr & \Rightarrow \log \frac{{\phi \left( v \right)}}{k} = \frac{{{x^2}}}{2}{\text{ or }}\phi \left( v \right) = k{e^{\frac{{{x^2}}}{2}}} \cr & \Rightarrow \phi \left( {xy} \right) = k{e^{\frac{{{x^2}}}{2}}} \cr} $$

Given DE can be written as
$$\eqalign{ & \int {dy} = \int {\frac{1}{{\left( {\sqrt {4 + \sqrt {9 + \sqrt x } } } \right)\left( {\sqrt {9 + \sqrt x } } \right)8\sqrt x }}dx} \cr & {\text{Putting }}\sqrt {4 + \sqrt {9 + \sqrt x } } = t,\,{\text{We get}} \cr & \frac{1}{{2\sqrt {4 + \sqrt {9 + \sqrt x } } .\,2\sqrt {9 + \sqrt x } .\,2\sqrt x }}dx = dt \cr & \therefore \int {dy = \int {dt\,\,\,\,\,\,\,\,\, \Rightarrow y = t + c} } \cr & {\text{or}},y = \sqrt {4 + \sqrt {9 + \sqrt x } } + C \cr & y\left( 0 \right) = \sqrt 7 \,\,\, \Rightarrow C = 0 \cr & \therefore y = \sqrt {4 + \sqrt {9 + \sqrt x } } \cr & \therefore y\left( {256} \right) = 3 \cr} $$

$$\eqalign{ & \sin \,x\frac{{dy}}{{dx}} + 2y\,\cos \,x = 1 \cr & \Rightarrow \frac{{dy}}{{dx}} + 2y\,\cot \,x = {\text{cosec}}\,x \cr & {\text{Here, }}P = 2\,\cot \,x,\,Q = {\text{cosec}}\,x \cr & {\text{Integrating factor }}I = {e^{\int {P\,dx} }} \cr & = {e^{\int {2\,\cot \,x\,dx} }} = {e^{2\int {\frac{{\cos \,x}}{{\sin \,x}}dx} }} \cr & = {e^{2\,\ln \,\sin \,x}}{\text{ Integration of }}\int {\cot \,x = \ln \left| {\sin \,x} \right| + C} \cr & = {\sin ^2}x \cr} $$

$$\eqalign{ & d\left( {xy} \right) = \frac{{f\left( {xy} \right)}}{{f'\left( {xy} \right)}}x\,dx \cr & {\text{or }}\frac{{f'\left( {xy} \right)}}{{f\left( {xy} \right)}}d\left( {xy} \right) = x\,dx\,\,\, \Rightarrow \int {\frac{{f'\left( {xy} \right)}}{{f\left( {xy} \right)}}d\left( {xy} \right)} = \int {x\,dx} \cr & {\text{or }}\log \left\{ {f\left( {xy} \right)} \right\} = \frac{{{x^2}}}{2} + c \cr & \therefore f\left( {xy} \right) = {e^{\frac{{{x^2}}}{2} + c}} = {e^c}.{e^{\frac{{{x^2}}}{2}}} = k{e^{\frac{{{x^2}}}{2}}} \cr} $$

$$\eqalign{ & {\text{We have,}}\,\,f\left( \theta \right) = \frac{d}{{d\theta }}\int\limits_0^\theta {\frac{{dx}}{{1 - \cos \,\theta \,\cos \,x}}} \cr & \Rightarrow f\left( \theta \right) = \frac{1}{{1 - {{\cos }^2}\theta }} \cr & \Rightarrow f\left( \theta \right) = {\text{cose}}{{\text{c}}^2}\theta \,\,\,\,\,\,\left( {{\text{using Leibnitz's Rule}}} \right) \cr & \Rightarrow \frac{{df\left( \theta \right)}}{{d\theta }} = - 2\,{\text{cose}}{{\text{c}}^2}\theta \,\cot \,\theta \cr & \Rightarrow \frac{{df\left( \theta \right)}}{{d\theta }} + 2\,f\left( \theta \right)\cot \,\theta = 0 \cr} $$