\(sin\left( \frac { dy }{ dx } \right) =a\)
\(\Rightarrow \frac { dy }{ dx } ={ sin }^{ -1 }(a)\)
\(\Rightarrow dy={ sin }^{ -1 }(a)dx\)
\(\Rightarrow \int { dy } ={ sin }^{ -1 }(a)\int { dx } \)
Taking Integration on both sides, we get
\(\Rightarrow \int { dy } ={ sin }^{ -1 }(a)\int { dx } \)
\(\Rightarrow y={ sin }^{ -1 }(a)x+c ...(1)\)
Initial condition:
Since y(0) = 1 we get,
1 = sin^-1(a)(0) +C
0 + C ⇒ C = 1
equation (1) ⇒ y = sin^-1(a) x + 1
y^-1 = sin^-1(a) + x
\(\Rightarrow \frac { y-1 }{ x } =sin(a)\Rightarrow sin\left( \frac { y-1 }{ x } \right) =a\)

The equation of the family of all ellipses having foci on the x -axis and centre at the origin is given by \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1,a>b\).
where a and b are arbitrary constants.
Differentiating equation (1) with respect to x, we get
\(\frac { 2x }{ { a }^{ 2 } } +\frac { 2y }{ { b }^{ 2 } } \frac { dy }{ dx } =0\quad \Rightarrow \frac { x }{ { a }^{ 2 } } +\frac { y }{ { b }^{ 2 } } \frac { dy }{ dx } =0\)
Differentiating equation (2) with respect to x, we get 
\(\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] =0\Rightarrow \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] \)
Substituting the value of \(\frac{1}{a^2}\) in equation (2) and simplifying, we get \(-\frac { 1 }{ { b }^{ 2 } } \left[ y\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] x+\frac { v }{ { b }^{ 2 } } \frac { dy }{ dx } =0\Rightarrow xy\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 2 }-y\frac { dy }{ dx } =0\) which is the required differential equation

Given y = ae^-3x+ b   ........(1)
Differentiating cquation (1) w.r.t 'x', we get
\(\frac { d y }{ d x } =ae^{ -3x }(-3)+0\)
\(\frac { d y }{ d{ x } } =ae^{ -3x }(-3) \)
Again differentiating, we get
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } ={ ae^{-3 x } }(+9)\)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } =-\frac{1}{3}\frac{dy}{dx}\times 9\)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } = {-3}\frac{dy}{dx} \)
\(\Rightarrow \frac { d^{ 2 }y }{ d{ x }^{ 2 } } +3\frac{dy}{dx} \) = 0 
Therefore, y = ae^-3x + b is a solution of the given differential equation.

given equation of family of curves is
y = Ae^8x + Be^-8x ..(1)
where A & B are arbitrary constants. Differentiating cquation (1) twice successively (because we have two arbitrary constant), we get
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } \\ \\ \) = 8Ae^8x + 8Be^-8x
Differentiating again with respect to 'x' we get,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64Ae^8x + 64Be^-8x
= 64(Ae^8x + Be^-8x)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = 64 y [using (1)]
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) - 64 y = 0
Which is the required differential equation.

Equation of family of parabolas with axis as y axis is given by,
(x-0) = 4a(y-k) .... (1)

Given: Vertex at (0, - 1).
Putting k = -1 in (1), we get
⇒ x² = \(\pm\)4a(y + 1) ....(2)
Differentiating with respect to 'x'  
2x = \(\pm\)4a\(\left( \frac { dy }{ dx } \right) \) ....(3)
⇒ 4a = \(\frac { 2x }{ \frac { dy }{ dx } } \) 
\(\frac{x^2}{2x} = \frac{y+1}{\frac{dt}{dx}}\)
ie) \(x \frac{dy}{dx}-2(y+1) =0\)
This is the required differential equation.

Given equation of curve is
xy = ae^x + be^−x + x² ...(1)
where a &b are aribitrary constant. differentiate equation (1) twice successively, because we have two arbitray constant.
\(x \frac{d y}{d x}+y(1)=a \mathrm{e}^{x}-\mathrm{be}^{-x}+2 x\) ...(2)
\( x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(1)+\frac{d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \)
\( x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}=\mathrm{ae}^{x}+\mathrm{be}^{-x}+2 \) ...(3)
From (1), we get \(x y-x^{2}=\mathrm{ae}^{x}+\mathrm{be}^{-x}\) ...(4)
Substituting equation (a) in (3), we get
\(\therefore x \frac{d^{2} y}{d x^{2}}+\frac{2 d y}{d x}-x y+x^{2}-2=0\) which is the required differential equaiton. 

Let V be the velocity and the retardation (negative acceleration) be -\(\frac{dv}{dt}\)
Given \(\frac{dv}{dt}\) = -V
Separating the variables,
\(\frac{dv}{v}=-dt\)
\(\Rightarrow \int { \frac { dv }{ v } } =-\int { dt } \)
\(\Rightarrow log\quad v=-t+logC\)
\(\Rightarrow logv-logC=-t\)
\(\Rightarrow log\left( \frac { v }{ { C }_{ v } } \right) =-t\)
\(\Rightarrow ={ e }^{ -t }\)
\(\Rightarrow \frac { v }{ { C }_{ v } } ={ Ce }^{ -t }...(1)\)
Given when t = 0, v = m/sec
\(\therefore\) (1) become 10 = C_e0 \(\Rightarrow\) C = 10
\(\therefore\) (1)  v = 10e^-t
When t = 2, v = 10e^-2
\(\Rightarrow v=\frac { 10 }{ { e }^{ 2 } } \)

\(x\frac { dy }{ dx } +2y=x^2logx\)
Dividing by x we get,
\(\frac { dy }{ dx } +\frac { 2 }{ x } y=xlogx\)
This is a linear differential equation 
\(\therefore P=\frac { 2 }{ x } ;Q=logx\)
\(\int { pdx } =\int { \frac { 2 }{ x } } dx=2logx=logx^2\)
\(\therefore I.F={ e }^{ \int { pdx } }={ e }^{ log\ x ^2}=x ^2\)
\(\therefore\) The solution is \({ e }^{ \int { u\ dv} }=uv-\int { vdu}\)
\(u=log\ x;dv=x^3\)
\(du=\frac { 1 }{ x }dx;v=\frac { { x }^{ 4} }{ 4} \)
\({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { xlogx.({ x }^{ 2 })dx } \)
\(\Rightarrow { x }^{ 2 }y=\int { { x }^{ 3 } } log\quad xdx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\int { \frac { { x }^{ 4 } }{ 4 } .\frac { 1 }{ x } } dx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { 1 }{ 4 } \int { { x }^{ 3 }dx } \)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { { x }^{ 4 } }{ 16 } +c\)

⇒ tan y\(\frac{dy}{dx}\) = cos (x+y) + cos (x-y)
⇒ tan y\(\frac{dy}{dx}\) = cos x cos y- sin x sin y + cos x cos y + sin x siny
[∵ cos (A+B) = cos A cos B- sin A sin B cos (A- B) = cos Acos B + sin A sin B]
= 2 cosx cosy
\(\Rightarrow \frac { tan\ y }{ cos\ y } dy=2\ cos\ x\ dx\)
Taking integration on both sides, we get
\(\Rightarrow \int { tan\ y\ sec\ y\ dy=2\int { cos\ x\ dx } } \)
\(\\ \\ \Rightarrow \ sec\ y=2sin\ x+c\)

⇒ (e^y+1) cos x dx + e^y sin x dy = 0
 \(\frac{\cos x d x}{\sin x}   =-\frac{e^y}{e^y+1} d y  \)
\(-\frac{e^y d y}{e^y+1}   =\cot x \mathrm{~d} x   \)
Take \( \mathrm{t}   =\mathrm{e}^y+1   \)
\(\mathrm{dt}   =\mathrm{e}^y \mathrm{dy}\)
Taking integration on both sides and substitute t and dt value, we get
 \(-\int \frac{d t}{t}=\int \cot x d x\)
-log t = log sin x + C₁
-log (e^y+1) = log(sin x) + C₁
log (sin x) = log(e^y+1) = log C
log (sin x) + log(e^y+1) = -C₁ = log C
log sin x(e^y+1) = log C
sin x (e^y+1) = C