\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +\int { ydx } ={ x }^{ 3 }\)
The given differential equation is
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +\int { ydx } ={ x }^{ 3 }.\)
Differentiating again with respect to 'x' we get.
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y=3{ x }^{ 2 }\)
The highest derivative is 3 and its power is 1.
∴ Order is 3 and degree is 1.

\({ x }^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +{ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=0\)
The given differential equation is
\({ x }^{ 2 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) \) = -\({ \left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }\)
= - \(\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides,
\({ x }^{ 4 }\left( \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) =1+{ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
The highest derivative is 2 and its power is 2.
∴ Order 2, degree 2.

The equation of the family of ellipses having centre at the origin & foci on the y-axis, is given
\(\frac { { x }^{ 2 } }{ { b }^{ 2 } } +\frac { { y }^{ 2 } }{ { a }^{ 2 } } =1\) ...(1)
where b >a & a, b are the parameters or a,b are arbitrary constant.
Differentiating equation (1) twice successively, because we have two arbitrary constant) we get
\( \frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \frac{d y}{d x} =0 \)
\(2\left(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}\right) =0 \)
\(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x} =0\) ............(2)
Again differentiating equation (2)
\(\frac{1}{a^{2}}+\frac{y}{b^{2}} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \frac{d y}{d x b^{2}}=0\)
\(\frac{1}{a^{2}}+\frac{y}{b^{2}} \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2} \frac{1}{b^{2}}=0\)
multiply by x
\(\frac{x}{a^{2}}+\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2} \frac{x}{b^{2}}=0\) .........(3) 
Equation (3)-(2)
\( \frac{x}{a^{2}}+\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2}\left(\frac{x}{b^{2}}\right) -\left(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}\right) =0 \)
\(\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2} \frac{x}{b^{2}}-\frac{y}{b^{2}} \frac{d y}{d x} =0 \)
Taking \(\frac{1}{b^{2}}\) outside, we get
\( \frac{1}{b^{2}}\left[x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}\right]=0 \\ x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0 \)
is the required differential equation.

Given circle equation be x² y² = r²
Let y = mx + c be the family of lines which touches the circle.
The condition for y = mx + c be all straight lines which towards the given circle x² y² = r² (1 + m²)
\(c=\sqrt { { 1+m }^{ 2 } } \)  
Hence, equation of tangent to the circle is .......(1)
y = mx + r\(\sqrt { { 1+m }^{ 2 } } \) .......(1) 
Differentiating with respect to x,
\(\frac { dy }{ dx } =m\quad ...(2)\)
Substituting 'm' in (1) we get,
\(y= \left( \frac { dy }{ dx } \right) \times x \pm r\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
\(y-x\left( \frac { dy }{ dx } \right) =\pm r\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } \)
Squaring both sides we get,
\(\Rightarrow { \left[ y-x\left( \frac { dy }{ dx } \right) \right] }^{ 2 }={ r }^{ 2 }\left[ 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] \)
This is the differential equation of all straight lines touching the circle x² y² = r²

General equation of a straight line in a plane is ax + by = 1  .....(1)
Since, the lines are non - horizontal, a ≠ 0 
Hence, differentiating with respect to y, equation (1)
a \(\\ \frac { dx }{ dy }+b =0\)
Differentiating again with respect to 'y'; we get,
\((a) \ \frac { { d }^{ 2 }x }{ d{ y }^{ 2 } } =0\Rightarrow \frac { { d }^{ 2 }x }{ d{ y }^{ 2 } } =0\quad [\because a\neq 0]\)
This is the differential equation of all non-horizontal lines in a plane.

Let x(t) denote the amount of salt in the tank at time t.
Its rate of change is
\(\frac{dx}{dt}\) = inflow rate - outflow rate
Now, 2 gram time 3 litres per minutes is inflow rate = 6 grams of salt. (3 x 2 = 6)
The out flow of salt is \(\frac{3}{50}\) times x = \(\frac{3x}{50}\)
\(\therefore \frac { dx }{ dt } =6-\frac { 3x }{ 50 } =\frac { 300-3x }{ 50 } \)
\(=-\frac { 3(x-100) }{ 50 } \)
\(\Rightarrow \frac { dx }{ x-100 } =-\frac { 3 }{ 50 } dt\)
\(\Rightarrow \int { \frac { dx }{ x-100 } =-\frac { 3 }{ 50 } \int { dt } } \)
\(\Rightarrow log(x-100)=-\frac { 3 }{ 50 } t+logC\)
\(\\ \Rightarrow log(x-100)-logC=-\frac { 3 }{ 50 } t\)
\(\Rightarrow log\left( \frac { x-100 }{ C } \right) =-\frac { 3 }{ 50 } t\)
\(\Rightarrow \frac { x-100 }{ C } ={ e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow x-100={ C }_{ e }-\frac { 3t }{ 50 } \quad ...(1)\)
When t = 0, x = 0
[Since initial water was pure without any salt]
\(\Rightarrow\) 0-100 = Ce⁰ 
\(\Rightarrow\) C = -100
(1) becomes x-100 = -100\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow\) x = 100-100\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\)
\(\Rightarrow\) x = 100(1-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))
Hence the amount of salt in the tank at time t is x = 100(1-\(\\ \\ \\ \\ \\ \\ \\ { e }^{ -\frac { 3t }{ 50 } }\))

Let T represent the temperature of the boiling water and Tm represents the temperature of the kitchen.
By Newton's law of cooling
\(\Rightarrow \int { \frac { dT }{ T-{ T }_{ m } } =K\int { dt } } \)
\(\Rightarrow log(T-{ T }_{ m })=Kt+logC\)
\(\Rightarrow log(T-{ T }_{ m })-logC=Kt\)
\(\Rightarrow log\left( \frac { T-{ T }_{ m } }{ C } \right) =Kt\)
\(\Rightarrow T-{ T }_{ m }={ Ce }^{ Kt } ...(1)\)
when t=0,T=100
\(\therefore 100-{ T }_{ m }={ Ce }^{ 0 }\)
\(\Rightarrow C=100-{ T }_{ m }\)
\(\Rightarrow becomes,\ T-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ Kt }\)
Also when t = 5, T = 80
\(\therefore 80-{ T }_{ m }=(100-{ T }_{ m }){ e }^{ 5K }\)
\(\Rightarrow { e }^{ 5K }=\frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } ..(2)\)
When t = 10, T = 65
(2) \(\Rightarrow\) 65 - T = (100-T_m)e^10K
= (100-T_m)(e^5K)²
\(=(100-{ T }_{ m }){ \left( \frac { 80-{ T }_{ m } }{ 100-{ T }_{ m } } \right) }^{ 2 }\)
[using(2)]
\(\Rightarrow 65-{ T }_{ m }=\frac { { (80-{ T }_{ m } })^{ 2 } }{ 100-{ T }_{ m } } \)
\(\Rightarrow\) 6500-65T_m-100T_m+T_m² = 6400+T_m²-160T_m
\(\Rightarrow\) 6500-6400 = 165Tm-160Tm
\(\Rightarrow\) 100 = 5Tm
\(\\ \Rightarrow { T }_{ m }=\frac { 100 }{ 5 } ={ 20 }^{ o }C\)
Hence the temperature of the kitchen is 20^oC

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 \({ e }^{ \int { u\ dv} }=uv-\int { vdu}\)
\(u=log\ x;dv=x^3\)
\(du=\frac { 1 }{ x }dx;v=\frac { { x }^{ 4} }{ 4} \)
\({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 2 }=\int { xlogx.({ x }^{ 2 })dx } \)
\(\Rightarrow { x }^{ 2 }y=\int { { x }^{ 3 } } log\quad xdx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\int { \frac { { x }^{ 4 } }{ 4 } .\frac { 1 }{ x } } dx\)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { 1 }{ 4 } \int { { x }^{ 3 }dx } \)
\(\Rightarrow { x }^{ 2 }y=\frac { { x }^{ 4 } }{ 4 } logx-\frac { { x }^{ 4 } }{ 16 } +c\)

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

\(\Rightarrow \frac { dy }{ dx } =\frac { y-x{ cos }^{ 2 }\left( \frac { y }{ x } \right) }{ x } ...(1)\)
This is a homogeneous differential equation
\(\therefore put\quad y=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dy }{ dx } \)
\(\therefore\)(1) becomes,
\(v+x\frac { dy }{ dx } =\frac { vx-x{ cos }^{ 2 }(v) }{ x } \)
\(\Rightarrow \frac { x(v-{ cos }^{ 2 }v) }{ x } =v-{ cos }^{ 2 }v\)
\(\Rightarrow x\frac { dv }{ dx } =v-{ cos }^{ 2 }(v)-v=-{ cos }^{ 2 }(v)\)
\(\Rightarrow \frac { dv }{ { cos }^{ 2 }v } =\frac { -dx }{ x } \Rightarrow { sec }^{ 2 }vdv=-\frac { dx }{ x } \)
Integrating,
\(\int { { sec }^{ 2 }vdv } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow tan\ v=-log\ x+log\ c\)
\(\Rightarrow tan\ v=log\left( \frac { c }{ x } \right) \)
\(\Rightarrow { e }^{ tan\ v }=\frac { c }{ x } \)
\(\Rightarrow c=c{ e }^{ tan\ v }\ \)
\(\Rightarrow c=x{ e }^{ tan\left( \frac { y }{ x } \right) }\ [\because v=\frac { y }{ x } ]\)

\(\left[ x+y\ cos\left( \frac { y }{ x } \right) \right] dx=x\ cos\left( \frac { y }{ x } \right) dy\)
\(\Rightarrow \frac { dy }{ dx } =\frac { x+ycos\left( \frac { y }{ x } \right) }{ xcos\left( \frac { y }{ x } \right) } ...(1)\)
This is homogeneous differential equation
\(\therefore put\ y=vx\)
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } ...(2)\)
Substituing (2) in (1)we get,
\(v+x\frac { dv }{ dx } =\frac { x+vxcosv }{ xcosv } \)
\(\frac { x(1+vcosv) }{ xcosv } =\frac { 1+vcosv }{ cosv } \)
\(\Rightarrow x\frac { dv }{ dx } =\frac { 1+vcosv }{ cosv } -v\)
\(=\frac { 1+vcosv-vcosv }{ cosv } \)
\(=\frac { 1 }{ cosv } \)
\(cosv\ dv=\frac { dx }{ x } \)
On integration, we obtain
\(\Rightarrow \int { cos\ v\ dv } =\int { \frac { dx }{ x } } \)
\(\Rightarrow sin\ v=log\ |x|+log\ |c|\)
\(\Rightarrow sin\left( \frac { y }{ x } \right) =log|cx|\)
\([\because y=vx\Rightarrow v=\frac { y }{ x } ]\)
which gives the required solution.

ydx + (1 + x²) tan^-1 xdy = 0
\(\mathrm{yd} x=-\left(1+x^2\right) \tan ^{-1} x \mathrm{dy} \)
\(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}=-\frac{d y}{y} \)
Take \(\mathrm{t}=\tan ^{-1} x \)
\(\mathrm{dt}=\frac{1}{1+x^2} d x\)
The equation can be written as
\(\frac{d t}{t}=-\frac{d y}{y}\)
Taking Integration on both sides, we get
\(\int \frac{d t}{t}=-\int \frac{d y}{y}\)
log t = - log y + log C
log (tan^-1 x) = -log y+ log C
log (tan^-1 x) + log y = log C
log y(tan^-1 x) = log c
y tan^-1 x = c

Given \(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
On separating the variables we get,
\(\frac { vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { g }{ { k }^{ 2 } } .dx\)
Multiplying by -2 both sides we get,
\(v\frac { dv }{ dx } =g\left( 1-\frac { { v }^{ 2 } }{ { k }^{ 2 } } \right) =g\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ { k }^{ 2 } } \right) \)
\(\frac { -2vdv }{ { k }^{ 2 }-{ v }^{ 2 } } =\frac { -2g }{ { k }^{ 2 } } dx\)
Taking integrating on both sides, we get
\(\int { \frac { -2v }{ { k }^{ 2 }-{ v }^{ 2 } } } =\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })=\frac { -2g }{ { k }^{ 2 } } .x+log\quad c\)
\(\Rightarrow log({ k }^{ 2 }-{ v }^{ 2 })-log\quad c=\frac { -2g }{ { k }^{ 2 } } .x\)
\(\Rightarrow log\left( \frac { { k }^{ 2 }-{ v }^{ 2 } }{ c } \right) -\frac { -2gx }{ { k }^{ 2 } } \)
\(\Rightarrow \frac { { k }^{ 2 }-{ v }^{ 2 } }{ e } ={ e }^{ -\frac { -2gx }{ { k }^{ 2 } } }\)
\(\Rightarrow { k }^{ 2 }-{ v }^{ 2 }{ ce }^{ \frac { -2gx }{ { k }^{ 2 } } }...(1)\)
Initial condition:
when v = 0, x = 0 we get
\( k^2-(0)^2 =C e^{\frac{-2g(0)}{k^2}} \)
\(k^2 =C e^0 \Rightarrow k^2=C \)
\((1) \Rightarrow k^2-v^2 =k^2 e^{\frac{-2 x^2}{k^2}} \)
\(k^2-k^2 e^{\frac{-2 s x}{k^1}} =\mathrm{v}^2 \)
\(k^2\left[1-e^{\frac{-2 s x}{k^2}}\right] =\mathrm{v}^2\)

Given equation is m \(\frac{dV}{dt}\) = F- kv  
The given equation can be written as
\(\frac { dv }{ F-kv } =\frac { dt }{ m } \)
Now Integrating, we get
\(\int { \frac { dv }{ F-kv } } =\int { \frac { dt }{ m } } \)
\(\int \frac{d V}{F-k V}=\int \frac{d t}{M}   \frac{\log (F-k V)}{-k}=\frac{t}{M}+C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}-k C_1 \\ \log [\mathrm{F}-\mathrm{kV}]=-\frac{k t}{M}+\log \mathrm{C} \\ \log [\mathrm{F}-\mathrm{kV}]-\log \mathrm{C}=-\frac{k t}{M} \\ \log \left(\frac{F-k V}{C}\right)=-\frac{k t}{M} \\ \frac{F-k V}{C}=e^{\frac{-k t}{M}} \\ \frac{F-k V}{e^{\frac{-t}{M}}}=\mathrm{C} \Rightarrow \mathrm{C}=e^{\frac{k t}{M}}(F-k V) \\\)
Initial condition:
Given V = 0 when t = 0
\(\mathrm{C}=e^{\frac{k(0)}{M}}[\mathrm{~F}-\mathrm{k}(0)] \\ =\mathrm{e}^0[\mathrm{~F}-0] \\ \mathrm{C}=\mathrm{F} \\ \therefore \mathrm{F}=(F-k V) e^{\frac{k t}{M}}\)