We have
$$\eqalign{ & \left( {2 + \sin \,x} \right)\frac{{dy}}{{dx}} + \left( {y + 1} \right)\cos \,x = 0 \cr & \Rightarrow \frac{d}{{dx}}\left( {2 + \sin \,x} \right)\left( {y + 1} \right) = 0 \cr} $$
On integrating, we get
$$\eqalign{ & \left( {2 + \sin \,x} \right)\left( {y + 1} \right) = C \cr & {\text{At }}x = 0,y = 1\,\,{\text{we}}\,\,{\text{have}} \cr & \left( {2 + \sin \,0} \right)\left( {1 + 1} \right) = C \cr & \Rightarrow C = 4 \cr & \Rightarrow y + 1 = \frac{4}{{2 + \sin \,x}} \cr & y = \frac{4}{{2 + \sin \,x}} - 1 \cr & {\text{Now }}y\left( {\frac{\pi }{2}} \right) = \frac{4}{{2 + \sin \,\frac{\pi }{2}}} - 1 \cr & = \frac{4}{3} - 1 \cr & = \frac{1}{3} \cr} $$

$$\eqalign{ & A{x^2} + B{y^2} = 1\,.....(1) \cr & Ax + By\frac{{dy}}{{dx}} = 0\,.....(2) \cr & A + By\frac{{{d^2}y}}{{d{x^2}}} + B{\left( {\frac{{dy}}{{dx}}} \right)^2} = 0\,.....(3) \cr & {\text{From}}\,\,(2)\,\,{\text{and}}\,\,(3) \cr & x\left\{ { - By\frac{{{d^2}y}}{{d{x^2}}} + B{{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right\} + By\frac{{dy}}{{dx}} = 0 \cr} $$
Dividing both sides by $$-B,$$  we get
$$ \Rightarrow xy\frac{{{d^2}y}}{{d{x^2}}} + x{\left( {\frac{{dy}}{{dx}}} \right)^2} - y\frac{{dy}}{{dx}} = 0$$
Which is a DE of order 2 and degree 1.

$$\eqalign{ & {\text{We have }}y = {c_1}{e^{{c_2}x}} \cr & \Rightarrow y' = {c_1}{c_2}\,{e^{{c_2}x}} = {c_2}y \cr & \Rightarrow \frac{{y'}}{y} = {c_2} \cr & \Rightarrow \frac{{y''y - {{\left( {y'} \right)}^2}}}{{{y^2}}} = 0 \cr & \Rightarrow y''y = {\left( {y'} \right)^2} \cr} $$

$$\eqalign{ & \frac{x}{{c - 1}} + \frac{y}{{c + 1}} = 1......\left( 1 \right) \cr & \Rightarrow \frac{1}{{c - 1}} + \frac{{y'}}{{c + 1}} = 0......\left( 2 \right) \cr & \Rightarrow \frac{{y'}}{1} = \frac{{c + 1}}{{1 - c}} \cr & \Rightarrow \frac{{y' - 1}}{{y' + 1}} = c \cr & {\text{Put the value of }}c{\text{ in equation}}\left( 1 \right) \cr & \Rightarrow \frac{x}{{\frac{{y' - 1}}{{y' + 1}} - 1}} + \frac{y}{{\frac{{y' - 1}}{{y' + 1}} + 1}} = 1 \cr & \Rightarrow \frac{{x\left( {y' + 1} \right)}}{{ - 2}} + \frac{{y\left( {y' + 1} \right)}}{{2y'}} = 1 \cr & \Rightarrow \frac{{\left( {y' + 1} \right)}}{2}\left( {\frac{y}{{y'}} - x} \right) = 1 \cr & \Rightarrow \left( {1 + \frac{{dy}}{{dx}}} \right)\left( {y - x\frac{{dy}}{{dx}}} \right) = 2\frac{{dy}}{{dx}} \cr} $$

$$\eqalign{ & {\text{Differentiate }}xy\left( x \right) = {x^2}y'\left( x \right) + 2xy\left( x \right) \cr & {\text{or }}xy\left( x \right) + {x^2}y'\left( x \right) = 0 \cr & {\text{or }}x\frac{{dy}}{{dx}} + y = 0 \cr & {\text{or }}\ln \,y + \ln \,x = \ln \,c \cr & {\text{or }}xy = c \cr} $$

The differential equation is $$\frac{{{d^2}y}}{{d{x^2}}} = \cos \,nx$$
Integrating we get
$$\frac{{dy}}{{dx}} = \frac{{\sin \,nx}}{n} + C.....\left( {\text{i}} \right)$$
Integrating again
$$\eqalign{ & y = - \frac{{\cos \,nx}}{{{n^2}}} + Cx + D \cr & \Rightarrow {n^2}y + \cos \,nx = {n^2}\left( {Cx + D} \right) \cr} $$

The given differential equation is
$$\eqalign{ & \frac{{dy}}{{dt}} - \frac{t}{{1 + t}}y = \frac{1}{{1 + t}} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{ - \int {\frac{t}{{1\, + \,t}}\,dt} }} = {e^{ - \int {\left( {1\, - \,\frac{t}{{1\, + \,t}}} \right)\,dt} }} = {e^{ - \left( {t\, - \,\log \,\left( {1\, + \,t} \right)} \right)}} \cr & = {e^{ - t}}.{e^{\log \,\left( {1\, + \,t} \right)}} = \left( {1 + t} \right){e^{ - \,t}} \cr} $$
$$\therefore $$ Solution is
$$\eqalign{ & y.{e^{ - t}}\left( {1 + t} \right) = \int {\frac{1}{{\left( {1 + t} \right)}}{e^{ - t}}\left( {1 + t} \right)dt + C} \cr & \Rightarrow y.{e^{ - t}}\left( {1 + t} \right) = - {e^{ - t}} + C \cr & \Rightarrow y = - \frac{1}{{1 + t}} + \frac{{C{e^t}}}{{1 + t}} \cr & {\text{Given that }}y\left( 0 \right) = - 1 \cr & \Rightarrow - 1 = - 1 + C \cr & \Rightarrow C = 0 \cr & \therefore y = \frac{1}{{1 + t}} \cr & \therefore y\left( 1 \right) = - \frac{1}{{1 + 1}} = - \frac{1}{2} \cr} $$

$$\eqalign{ & y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n} \cr & \frac{{dy}}{{dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}\left( {1 + \frac{1}{2}{{\left( {1 + {x^2}} \right)}^{ - \frac{1}{2}}},2x} \right)\,; \cr & \frac{{dy}}{{dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}\frac{{\left( {\sqrt {1 + {x^2}} + x} \right)}}{{\sqrt {1 + {x^2}} }} = \frac{{n{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}}}{{\sqrt {1 + {x^2}} }} \cr & {\text{or }}\sqrt {1 + {x^2}} \frac{{dy}}{{dx}} = ny \cr & {\text{or }}\sqrt {1 + {x^2}} {y_1} = ny\,\,\,\,\,\,\,\,\left( {{y_1} = \frac{{dy}}{{dx}}} \right) \cr & {\text{Squaring, }}\left( {1 + {x^2}} \right)y_1^2 = {n^2}{y^2} \cr & {\text{Differentiating,}} \cr & \left( {1 + {x^2}} \right)2{y_1}{y_2} + y_1^2.2x = {n^2}.2y{y_1} \cr & \left( {{\text{Here }}{y_2} = \frac{{{d^2}y}}{{d{x^2}}}} \right) \cr & {\text{or }}\left( {1 + {x^2}} \right){y_2} + x{y_1} = {n^2}y \cr} $$

$$\eqalign{ & \sqrt {1 + {x^2}} + \sqrt {1 + {y^2}} = \lambda \left( {x\sqrt {1 + {y^2}} - y\sqrt {1 + {x^2}} } \right) \cr & \Rightarrow \sqrt {1 + {x^2}} \left( {1 + \lambda y} \right) = \sqrt {1 + {y^2}} \left( {\lambda x - 1} \right) \cr & \Rightarrow \frac{{\sqrt {1 + {x^2}} }}{{\sqrt {1 + {y^2}} }} = \frac{{\lambda x - 1}}{{\lambda y + 1}} \cr & \Rightarrow \frac{{{x^2} + 1}}{{{y^2} + 1}} = \frac{{{\lambda ^2}{x^2} - 2\lambda x + 1}}{{{\lambda ^2}{y^2} + 2\lambda y + 1}} \cr & \Rightarrow \left( {{y^2} + 1} \right)\left( {{\lambda ^2}{x^2} - 2\lambda x + 1} \right) = \left( {{x^2} + 1} \right)\left( {{\lambda ^2}{y^2} + 2\lambda y + 1} \right) \cr & \Rightarrow {\lambda ^2}{x^2}{y^2} - 2\lambda x{y^2} + {y^2} + {\lambda ^2}{x^2} - 2\lambda x + 1 = {\lambda ^2}{x^2}{y^2} + 2\lambda {x^2}y + {x^2} + {\lambda ^2}{y^2} + 2\lambda y + 1 \cr & \Rightarrow {\lambda ^2}\left( {{x^2} - {y^2}} \right) - 2\lambda \left( {x{y^2} + {x^2}y + x + y} \right) = 0 \cr & \Rightarrow {\lambda ^2}\left( {x + y} \right)\left( {x - y} \right) - 2\lambda \left[ {xy\left( {x + y} \right) + \left( {x + y} \right)} \right] = 0 \cr & \Rightarrow \lambda \left( {x + y} \right)\left[ {\lambda \left( {x - y} \right) - 2xy - 2} \right] = 0 \cr & \Rightarrow \left( {x + y} \right)\left[ {\lambda \left( {x - y} \right) - 2xy - 2} \right] = 0 \cr & \Rightarrow \lambda \left( {x - y} \right) - 2xy - 2 = 0 \cr & \Rightarrow \frac{{2xy + 2}}{{x - y}} = \lambda \cr & \Rightarrow \frac{{xy + 1}}{{x - y}} = \frac{\lambda }{2} \cr & \Rightarrow \frac{{\left( {x\frac{{dy}}{{dx}} + y} \right)\left( {x - y} \right) - \left( {xy + 1} \right)\left( {1 - \frac{{dy}}{{dx}}} \right)}}{{{{\left( {x - y} \right)}^2}}} = 1 \cr} $$
This is the first order differential equation and clearly degree of $$\frac{{dy}}{{dx}}$$  is 1. Hence degree of the differential equation is 1.

Consider the given differential equation
$$\eqalign{ & \ln \left( {\frac{{dy}}{{dx}}} \right) + x = 0 \cr & \Rightarrow \ln \left( {\frac{{dy}}{{dx}}} \right) = - x \cr & \Rightarrow \frac{{dy}}{{dx}} = {e^{ - x}} \cr} $$
on separating the variables, we get $$dy = {e^{ - x}}dx,$$
on integrating both sides, we get
$$\eqalign{ & \int {dy = \int {{e^{ - x}}dx} } \cr & \Rightarrow y = \frac{{{e^{ - x}}}}{{ - 1}} + C \cr & \Rightarrow y = - {e^{ - x}} + C \cr & \Rightarrow y + {e^{ - x}} = C \cr} $$