Given $$y = {e^{4x}} + 2{e^{ - x}}$$
Differentiating we get
$$\eqalign{ & \frac{{dy}}{{dx}} = 4{e^{4x}} - 2{e^{ - x}} \cr & \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = 16{e^{4x}} + 2{e^{ - x}} \cr & \Rightarrow \frac{{{d^3}y}}{{d{x^3}}} = 64{e^{4x}} - 2{e^{ - x}} \cr} $$
Putting these values in $$\frac{{{d^3}y}}{{d{x^3}}} + A\frac{{dy}}{{dx}} + By = 0$$
We have
$$\eqalign{ & \left( {64 + 4A + B} \right){e^{4x}} + \left( { - 2 - 2A + 2B} \right){e^{ - x}} = 0 \cr & \Rightarrow 64 + 4A + B = 0,\,\, - 2 - 2A + 2B = 0 \cr} $$
Solving these equations, we get $$A = - 13,\,\,B = - 12$$

$$\eqalign{ & \frac{{dV\left( t \right)}}{{dt}} = - k\left( {T - t} \right)\,\,\,\,\, \Rightarrow \int {dVt} = - k\int {\left( {T - t} \right)dt} \cr & V\left( t \right) = \frac{{k{{\left( {T - t} \right)}^2}}}{2} + c \cr & V\left( 0 \right) = I\,\, \Rightarrow I = \frac{{K{T^2}}}{2} + C\,\,\, \Rightarrow C = I - \frac{{K{T^2}}}{2} \cr & \therefore \,V\left( T \right) = 0 + C = I - \frac{{K{T^2}}}{2} \cr} $$

$$\eqalign{ & \sin \left( {\frac{{dy}}{{dx}}} \right) - a = 0 \cr & \sin \left( {\frac{{dy}}{{dx}}} \right) = a \cr & \Rightarrow \frac{{dy}}{{dx}} = {\sin ^{ - 1}}a\,;\,dy = {\sin ^{ - 1}}a\,dx \cr} $$
Now integrating both sides,
$$\eqalign{ & \int {dy} = \int {{{\sin }^{ - 1}}a\,dx} \cr & y = x\,{\sin ^{ - 1}}a + c \cr} $$

The given equation is
$$\eqalign{ & ydx + {y^2}dy = xdy;\,\,x \in R,\,y > 0,\,y\left( 1 \right) = 1\, \cr & \Rightarrow \frac{{ydx - xdy}}{{{y^2}}} + dy = 0 \cr & \Rightarrow \frac{d}{{dx}}\left( {\frac{x}{y}} \right) + dy = 0 \cr} $$
On integration, we get
$$\eqalign{ & \frac{x}{y} + y = C \cr & y\left( 1 \right) = 1\,\,\,\, \Rightarrow 1 + 1 = C\,\, \Rightarrow C = 2 \cr & \therefore \,\frac{x}{y} + y = 2 \cr} $$
Now to find $$y\left( { - 3} \right),$$   putting $$x=-3$$   in above equation, we get
$$\eqalign{ & - \frac{3}{y} + y = 2 \cr & \Rightarrow {y^2} - 2y - 3 = 0 \cr & \Rightarrow y = 3,\,\, - 1 \cr} $$
But given that $$y>0,$$
$$\therefore y = 3$$

$$\eqalign{ & \cos \,x\,dy = y\left( {\sin \,x - y} \right)dx \cr & \frac{{dy}}{{dx}} = y\,\tan \,x - {y^2}\,\sec \,x \cr & \frac{1}{{{y^2}}}\frac{{dy}}{{dx}} - \frac{1}{y}\tan \,x = - \sec \,x\,.....(i) \cr & Let\,\,\frac{1}{y} = t\,\,\,\,\, \Rightarrow - \frac{1}{{{y^2}}}\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} \cr & {\text{From equation }}(i) \cr & - \frac{{dt}}{{dx}} - t\,\tan \,x = - \sec \,x \cr & \Rightarrow \frac{{dt}}{{dx}} + \left( {\tan \,x} \right)t = \sec \,x \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\tan \,x\,dx} }} = {\left( e \right)^{\log \,\left| {\sec \,x} \right|}}\sec \,x \cr & {\text{Solution:}} \cr & t\left( {{\text{I}}{\text{.F}}{\text{.}}} \right) = \int {\left( {{\text{I}}{\text{.F}}{\text{.}}} \right)\sec \,x\,dx\, \Rightarrow \frac{1}{y}\sec \,x = \tan \,x + c} \cr} $$

On removal of the radical, $$1 + {\left( {\frac{{dy}}{{dx}}} \right)^2} = {x^4}.$$    So, degree $$=2.$$

Consider the given differential equation the
$$\eqalign{ & \sin \,xdy + y\,\cos \,xdx = 4xdx \cr & \Rightarrow d\left( {y.\sin \,x} \right) = 4xdx \cr} $$
Integrate both sides
$$\eqalign{ & \Rightarrow y.\sin \,x = 2{x^2} + C\,.....(1) \cr & \Rightarrow y\left( x \right) = \frac{{2{x^2}}}{{\sin \,x}} + c\,.....(2) \cr} $$
$$\because $$ equation (2) passes through $$\left( {\frac{\pi }{2},\,\,0} \right)$$
$$ \Rightarrow 0 = \frac{{{\pi ^2}}}{2} + C\,\,\, \Rightarrow C = - \frac{{{\pi ^2}}}{2}$$
Now, put the value of $$C$$ in (1)
Then, $$y\,\sin \,x = 2{x^2} - \frac{{{\pi ^2}}}{2}$$     is the solution
$$\therefore y\left( {\frac{\pi }{6}} \right) = \left( {2.\frac{{{\pi ^2}}}{{36}} - \frac{{{\pi ^2}}}{2}} \right)2 = - \frac{{8{\pi ^2}}}{9}$$

The equation of the family is
$$\eqalign{ & {\left( {x - c} \right)^2} + {y^2} = {c^2}{\text{ or }}{x^2} + {y^2} - 2cx = 0 \cr & {\text{or }}\frac{{{x^2} + {y^2}}}{x} = 2c\,\,\,\,\,\,\,\,\, \Rightarrow \frac{{\left( {2x + 2y\frac{{dy}}{{dx}}} \right)x - \left( {{x^2} + {y^2}} \right).1}}{{{x^2}}} = 0 \cr} $$
So, degree $$=1$$  order $$=1$$

$$\eqalign{ & 3{e^x}\tan \,y\,dx + \left( {1 - {e^x}} \right){\sec ^2}y\,dy = 0 \cr & \Rightarrow \frac{{3{e^x}}}{{1 - {e^x}}}dx + \frac{{{{\sec }^2}y}}{{\tan \,y}}dy = 0\,\,{\text{Integrating we get}} \cr & \int {\frac{{3{e^x}}}{{1 - {e^x}}}dx} + \int {\frac{{{{\sec }^2}y}}{{\tan \,y}}dy} = D \cr & \Rightarrow - 3\ell n\left( {1 - {e^x}} \right) + \ell n\,\tan \,y = D \cr & \Rightarrow - \ell n{\left( {1 - {e^x}} \right)^3} + \ell n\,\tan \,y = D \cr & \Rightarrow \ell n\frac{{\tan \,y}}{{{{\left( {1 - {e^x}} \right)}^3}}} = \ell n\,C \cr & \Rightarrow \tan \,y = C{\left( {1 - {e^x}} \right)^3} \cr} $$

$$\eqalign{ & {\text{Since, }}x\frac{{dy}}{{dx}} + 2y = \,{x^2} \cr & \Rightarrow \frac{{dy}}{{dx}} + \frac{2}{x}y = x \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{\int {\frac{2}{x}\,dx} }} = {e^{2\,\ln \,x}} = {e^{\ln \,{x^2}}} = {x^2} \cr} $$
Solution of differential equation is:
$$\eqalign{ & y\,.\,\% \,{x^2} = \int {x.{x^2}dx} \cr & y\,.\,\% \,{x^2} = \frac{{{x^4}}}{4} + C.....\left( 1 \right) \cr & \because y\left( a \right) = 1 \cr & \therefore C = \frac{3}{4} \cr} $$
Then, from equation (1)
$$\eqalign{ & y.\,\% \,{x^2} = \frac{{{x^4}}}{4} + \frac{3}{4} \cr & \therefore y = \frac{{{x^2}}}{4} + \frac{3}{{4{x^2}}} \cr & \therefore y\left( {\frac{1}{2}} \right) = \frac{1}{{16}} + 3 = \frac{{49}}{{16}} \cr} $$