order 2 ; degree 1

Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution

The equation of the family of parabolas is given by y² ax = 4, a is an arbitrary constant. ... (1)
Differentiating both sides of (1) with respect to x , we get 2y\(\frac{dy}{dx}=4a\Rightarrow a=\frac{y}{2}\frac{dy}{dx}\) 
Substituting the value of a in (1) and simplifying, we get \(\frac{dy}{dx}=\frac{y}{2x}\) as the required differential equation.

y''− 5y' + 6y = 0 ......(1)
Given y = e^mx  .....(2)
Differentiating cquation (2) w.r.t 'x', we get
\(\frac{dy}{dx} = em^x . m\)
To find the value of m:
Given y" - 5y' + 6y = 0 
e^mx . m² -5e^mx+ 6e^mx = 0
e^mx [m- 5m +6] = 0
m - 5m + 6 = 0
(m - 3) (m - 2) = 0
m = 3, 2

y = 2x² ; xy' = 2y
Consider y = ax² ...(1)
Differentiating with respect to 'x' we get,
⇒ \(\frac { dy }{ dx } = 2.2x\)  ...(2)
Multiply by x on both sides, we get,
x = \(\frac { dy }{ dx } = 2.2x\) 
y' = 2y is a given differential equation
Thus y = 2x² satislies the differential equation
xy' = 2y 
Hence y = 2x² is a solution of the differential equation xy' = 2y.

Let x represent the principal in the saving amount.
R = 8% and N = 1.
∴ Interest = \(\frac { PNR }{ 100 } =\frac { x\times 1\times 8 }{ 100 } =\frac { 2x }{ 25 } \)
∴ Given \(\frac { dx }{ dt } \) = interest + Rs. 400.
∴ \(\frac { dx }{ dt } =\frac { 2x }{ 25 } +400\)

prove

Given \(\frac { dy }{ dx } +\frac { { y }^{ 2 } }{ { x }^{ 2 } } =\frac { y }{ x } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { y }{ x } -\frac { { y }^{ 2 } }{ { x }^{ 2 } } ...(1)\)
This is a homogeneous differential equation
put y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) becomes,
\(v+x\frac { dv }{ dx } =\frac { vx }{ x } -\frac { { v }^{ 2 }{ x }^{ 2 } }{ { x }^{ 2 } } =v-{ v }^{ 2 }\)
\(x\frac { dv }{ dx } =v-{ v }^{ 2 }-v=-v\)
\(\therefore \frac { dv }{ { v }^{ 2 } } =\frac { -dx }{ x } \)
\(\Rightarrow \int { \frac { dv }{ { v }^{ 2 } } =-\int { \frac { dx }{ x } } } \)
\(\Rightarrow \frac { -1 }{ v } =-log\quad x+c\)
\(\Rightarrow \frac { -1 }{ \frac { y }{ x } } =-log\quad x+c\)
\(\Rightarrow \frac { -x }{ y } +logx=c\)

Separating the variables we get,
\(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
Taking Integration on both sides, we get
\(\int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{d x}{\sqrt{1-x^{2}}}\)
sin^-1y = sin^-1 x + c

Given that \((1+{ x }^{ 2 })\frac { dy }{ dx } =1+{ y }^{ 2 }\) ..(1)
The given equation is written in the variables separable form 
\(\frac { dy }{ { 1+y }^{ 2 } } =\frac { dx }{ { 1+x }^{ x } } \) ...(2)
Integrating both sides of (2), we get tan⁻¹ tan⁻¹x +C.
But tan^-1 y - tan^-1 x = tan^-1 \(\left( \frac { y-x }{ 1+xy } \right) .\) ...(4)
Using (4) in (3) leads to tan^-1 \(\left( \frac { y-x }{ 1+xy } \right)\) = C, which implies \(\frac { y-x }{ 1+xy } \) = tan C = a (say).
Thus, y − x = a(1+ xy) gives the required solution

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 the circles centre on r-axis & the circle is passing through the origin.
Let it be (r, 0) & its radius r.

Equation of the circle is
(x - a)² + (y - b)² = r²
(x - r)² + (y - 0)² = r²
⇒ x² - 2xr + r² + y² = r²
⇒ x² - 2xr + y² = 0 ...(1)
defferentiating equation (1) with respect to 'x' we get
⇒ 2x - 2r + 2y \(\frac { dy }{ dx } =0\)
⇒ 2x + 2y \(\frac { dy }{ dx } =2r\)
⇒ x + y \(\frac { dy }{ dx } =r\) ...(2)
Substituting r value in equation (1), we get
x² - 2x \(\left( x+y\frac { dy }{ dx } \right) +{ y }^{ 2 }=0\)
\(\Rightarrow \ { x }^{ 2 }-{ 2x }^{ 2 }-2xy\frac { dy }{ dx } +{ y }^{ 2 }=0\)
\(\Rightarrow \ { -x }^{ 2 }{ -2x }y\left( \frac { dy }{ dx } \right) { +y }^{ 2 }\)
Multiply by '-', we get
\(\Rightarrow \ { x }^{ 2 }{ +2x }y\left( \frac { dy }{ dx } \right) { -y }^{ 2 }\) which is the required differential equation.

Let T be the temperature of the coffee at time t
and T_m' the temperature of the kitchen.
By Newton's law of cooling
\(\frac { dT }{ dt } =K(T-{ T }_{ m })\)
\(\Rightarrow \frac { dT }{ dt } =K(T-70)\)
\(\Rightarrow \int { \frac { dT }{ T-70 } =K\int { dt } } \)
\(\Rightarrow log(T-70)=kt+logC\)
\(\Rightarrow log(T-70)-logC=Kt\)
\(\Rightarrow log\left( \frac { T-70 }{ C } \right) =Kt\)
\(\Rightarrow \frac { T-70 }{ C } ={ e }^{ Kt }\)
\(\Rightarrow T-70={ Ce }^{ Kt }...(1)\)
\(\\ When\ t=0,\ T={ 180 }^{ o }F\)
\(\therefore { 180 }^{ o }-{ 70 }^{ o }={ Ce }^{ 0 }\)
\(\Rightarrow C={ 11 }0^{ 0 }\)
\(\\ \therefore (1)\Rightarrow T-70=110{ e }^{ Kt } ..(2)\)
\(When\ t=0,T=160\)
\(\therefore 160-70=110{ e }^{ 10K }\)
\(90=110{ e }^{ 10K }\)
\(\Rightarrow { e }^{ 10K }=\frac { 9 }{ 11 } \)
\(\Rightarrow { e }^{ K }={ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } }...(3)\)
(i) when t = 15, (2) becomes,
\(\Rightarrow T-70=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 1 }{ 10 } \times 15 }\)
\(=110{ \left( \frac { 9 }{ 11 } \right) }^{ \frac { 3 }{ 2 } }\)
\(=110\times { \left( \frac { 9 }{ 11 } \right) }\left( \sqrt { \frac { 9 }{ 11 } } \right) \)
\(=110\times \frac { 9 }{ 11 } \times \frac { 3 }{ \sqrt { 11 } } \)
\(=\frac { 270 }{ \sqrt { 11 } } =\frac { 270 }{ 3.32 } =81.33\)
\(\Rightarrow\) T=81.33+70=151.3F
\(\therefore\) T = 151.3F
\(\therefore\) The temperature of the coffee at 10.15 am is 151.3F
(ii) when T = 130F, (2) becomes
T-70 = 110e^kt  ...(2)
\(\Rightarrow\) 130-70 = 110e^kt 
60 = 110e^kt 
e^kt  = \(\frac{6}{11}\)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 6 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 6 }{ 11 } \right) }{ log\left( \frac { 9 }{ 11 } \right) } \)
\(=\frac { log(0.545) }{ log(0.818) } =\frac { -0.264 }{ -0.087 } \)
= 3.34
t = 30.34min
T = 140F (2)becomes
140-70 = 110e^kt ...(2)
\(\Rightarrow 70={ 110e }^{ kt }\)
\({ e }^{ kt }=\frac { 7 }{ 11 } \)
\({ \left( \frac { 9 }{ 11 } \right) }^{ \frac { t }{ 10 } }=\frac { 7 }{ 11 } \)
\(\frac { t }{ 10 } =\frac { log\left( \frac { 7 }{ 11 } \right) }{ log\left( \frac { 7 }{ 11 } \right) } =\frac { -0.197 }{ -0.087 } \)
= 2.26
t = 22.6min
\(\therefore\) Between 10.22 min to 10.30 min, the woman should have drunk the coffee.

Let there be N radioactive nuclei in a sample at any time t and let N_o be the initial number of radioactive nuclei.
Then \(\frac{dN}{dt}\infty N\)
\(\Rightarrow \frac { dN }{ dt } =-\lambda N\)
Where \(\lambda>0\) is a constant
\(\Rightarrow \frac { dN }{ N } =-\lambda dt\)
\(\int { \frac { dN }{ N } } =-\lambda dt\)
\(\int { \frac { dN }{ N } =-\lambda \int { dt } } \)
\(\Rightarrow log\ N=-\lambda t+C\ ...(1)\)
\(T\quad t=0,\ we\ have\ N={ N }_{ 0 }\)
\(\therefore log{ N }_{ 0 }=0+C\)
\(\Rightarrow C=log{ N }_{ 0 }\)
\(\therefore\)(1) becomes, log N = -\(\lambda t+log{ N }_{ 0 }\)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } =-\lambda t\quad ...(2)\)
It is given that 10% of the original number of nuclei have undergone disintegration in a period of 100 years.
Whent= 100 \(N={ N }_{ 0 }-\frac { 10 }{ 100 } \times { N }_{ 0 }=\frac { { 9N }_{ 0 } }{ 10 } \)
Substituting in (2) we get
\(log\quad \frac { 9 }{ 10 } =-100\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 100 } log\frac { 9 }{ 10 } \)
Substituting in (2) we get,
\(log\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 100 } log\frac { 9 }{ 10 } \right) t\)
when t = 1000,
\(log\frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 100 } log\left( \frac { 9 }{ 10 } \right) \times 1000\)
\(=10log\left( \frac { 9 }{ 10 } \right) \)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } \times 100={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\times 100=\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \)
Hence, \(\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \%\) of radioactive nuclei will remain after 1000 years,

\(\frac { dy }{ dx } +\frac { { 3x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation 
\(\therefore P=\frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } \)
\(\therefore \int { pdx } =\int { \frac { { 3x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =log(1+{ x }^{ 3 })\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log(1+{ x }^{ 3 }) }=(1+{ x }^{ 3 })\)
\(\therefore\)The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { { sin }^{ 2 }x }{ 1+{ x }^{ 3 } } (1+{ x }^{ 3 })dx+c } \)
\(cos2x=1-2{ sin }^{ 2 }x\)
\(sin2x=\frac { 1-cos2x }{ 2 } =\int { { sin }^{ 2 }xdx+c } \)
\(\Rightarrow y(1+{ x }^{ 3 })=\int { \frac { 1-cos2x }{ 2 } } dx+c\)
\(\Rightarrow y(1+{ x }^{ 3 })=\frac { x }{ 2 } -\frac { sin2x }{ 4 } +c\)

The given differential equation can be written as
\(\frac{x \sin x}{x \sin x} \frac{d y}{d x}+\frac{(x \cos x+\sin x) y}{x \sin x} =\frac{\sin x}{x \sin x}\)  
\(\frac{d y}{d x}+\left(\frac{x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}\right) y  =\frac{1}{x}  \)
\(\frac{d y}{d x}+\left(\cot x+\frac{1}{x}\right) y  =\frac{1}{x}\)
This is of the form \(  \frac{d y}{d x}+P y=\mathrm{Q} \), where
\(\mathrm{P}=\cot x+\frac{1}{x} ; \mathrm{Q}=\frac{1}{x} \text {. }\)
Thus, the given differential equation is linear. 
\(  \text { I.F }=e^{\int h t s}=e^{\int(x+x+16 x d t}=e^{b \text { bedex+ive }} \)
\( =e^{\ln (\sin a)}=x \sin x\)
So, the solution of the given differential equatior is given by 
\(\mathrm{y} \times \mathrm{I} . \mathrm{F}   =\int(Q \times I . F) d x+c  \)
\(\mathrm{y}(x \sin x)   =\int \frac{1}{x} x \sin x d x+c  \)
\( =\int \sin x d x+c \)
\(\mathrm{y}(x \sin x)  =-\cos x+c\)
\(xy  \sin x+\cos x=\mathrm{c}\) is the required solution.

The given differential equation can be written as
\(\frac { dy }{ dx } +\left( \frac { -x }{ 1-{ x }^{ 2 } } \right) y=\frac { 1 }{ 1-{ x }^{ 2 } } \)
This is of the form \(\frac{dy}{dx}+Py = Q\)
where \( P=\frac { -x }{ 1-{ x }^{ 2 } } ;Q=\frac { 1 }{ 1-{ x }^{ 2 } } \)
Thus, the given differential equation is linear.
\(\int { pdx } =\int { \frac { -x }{ 1-{ x }^{ 2 } } dx } =\frac { 1 }{ 2 } \int { \frac { -2x }{ 1-{ x }^{ 2 } } dx } \)
\(=\frac { 1 }{ 2 } log(1-{ x }^{ 2 })=log\sqrt { 1-{ x }^{ 2 } } \)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log\sqrt { 1-{ x }^{ 2 } } }=\sqrt { 1-{ x }^{ 2 } } \)
So, the required solution is given by 
\({[\mathrm{y} \times \mathrm{I} . \mathrm{F}] }   =\int[Q \times I . F] d x+c \)
\(y \times \sqrt{1-x^2}   =\int \frac{1}{1-x^2} \sqrt{1-x^2} d x+c \)
\( =\int \frac{1}{\sqrt{1-x^2}} \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} d x+c \)
\( =\int \frac{1}{\sqrt{1-x^2}} d x+c \)
\(\div \sqrt{1-x^2}, \frac{y \sqrt{1-x^2}}{\sqrt{1-x^2}}   =\frac{\sin ^{-1} x}{\sqrt{1-x^2}}+\frac{c}{\sqrt{1-x^2}} \)
\(\mathrm{y}   =\frac{\sin ^{-1} x+c}{\sqrt{1-x^2}}+c\left(1-x^2\right)^{-1 / 2}\)
Which is a required solution.

The given equation can be written as \(\frac { dy }{ dx } -\frac{2}{y}x=y^2e^{-y}\).
This is a linear differential equation. Here \(P=-\frac { 2 }{ y } ;Q={ y }^{ 2 }{ e }^{ -y }\)
\(\int { pdy } =\int { -\frac { 2 }{ y } dy=-2log|y|=log{ |y| }^{ -2 } } =log\left( \frac { 1 }{ { y }^{ 2 } } \right) ,\)
Thus, \(I.F={ e }^{ \int { Pdy } }={ e }^{ log\left( \frac { 1 }{ { y }^{ 2 } } \right) }=\frac { 1 }{ { y }^{ 2 } } .\)
Hence the solution is \(x{ e }^{ \int { Pdy } }=\int { Q{ e }^{ \int { Pdy } }dy+C } \)
Thus, \(x\left( \frac { 1 }{ { y }^{ 2 } } \right) =\int { { y }^{ 2 }{ e }^{ -y } } \left( \frac { 1 }{ { y }^{ 2 } } \right) dy+C=\int { { e }^{ -y }dy+C } =-{ e }^{ -y }+C\)or x = −y²e^−y +Cy² is the required solution

Here, to make the coefficient of \(\frac { dy }{ dx } \) unity, divide both sides by (1+x³).
Then the equation is \(\frac { dy }{ dx } +\frac { { 6x }^{ 2 }y }{ 1+{ x }^{ 3 } } =\frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } \)
This is a linear differential equation in y.
Here, \(P=\frac { { 6x }^{ 2 } }{ 1+{ x }^{ 3 } } ;Q=\frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } \)
\(\int { Pdx } =\int { \frac { { x }^{ 2 } }{ 1+{ x }^{ 3 } } dx } =2log|1+{ x }^{ 3 }|=log{ |1+{ x }^{ 3 }| }^{ 2 }=log{ ({ 1+x }^{ 3 }) }^{ 2 }\)
Thus, I.F.\(={ e }^{ \int { Pdx } }={ e }^{ log{ ({ 1+x }^{ 3 }) }^{ 2 } }={ (1+{ x }^{ 3 }) }^{ 2 }\)
Hence the solution is \(ye^{ \int { Pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C. } \)
Thus is, \(y{ (1+{ x }^{ 2 }) }^{ 2 }=\int { \frac { 1+{ x }^{ 2 } }{ 1+{ x }^{ 3 } } } { (1+{ x }^{ 3 }) }^{ 2 }dx+C=\int { (1+{ x }^{ 2 }) } (1+{ x }^{ 3 })dx+C=\int { (1+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 5 }) } dx+C\) \(or\ y{ (1+{ x }^{ 3 }) }^{ 2 }=x+\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +\frac { x^{ 6 } }{ 6 } +C\)
\(and\quad y=\frac { 1 }{ { (1+{ x }^{ 3 }) }^{ 2 } } \left[ x+\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +\frac { x^{ 6 } }{ 6 } +C \right] \)is the required solution.

\(\left[ x+y\ cos\left( \frac { y }{ x } \right) \right] dx=x\ cos\left( \frac { y }{ x } \right) dy\)
\(\Rightarrow \frac { dy }{ dx } =\frac { x+ycos\left( \frac { y }{ x } \right) }{ xcos\left( \frac { y }{ x } \right) } ...(1)\)
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.

\(\Rightarrow \frac { dy }{ dx } -x\sqrt { 25-{ x }^{ 2 } } \)
\(\Rightarrow \frac { dy }{ dx } -x\sqrt { 25-{ x }^{ 2 } } \)dx
put 25 - x² = t²
⇒ -2x dx = 2t dt
x dx = -dt
Putting in (1),
⇒ x dx = \(\frac{-dt}{2}\)
dy = t \(\times\)(-t) dt
Integrating on both sides,
\( \int d y =-\int t^2 d t \)
\(y =\frac{-t^3}{3}+C_1 \Rightarrow 3 y+t^3=3 C_1 \)
\(3 y+\left(25-x^2\right)^{\frac{3}{2}} =C\)

By putting z = 4x + 2y −1, we have
z' = 4+2y' = 4+2\(\sqrt z\)
hence \(\frac { dz }{ 4+2\sqrt { z } } =dx\).
Integrating, \(\int { \frac { dz }{ 4+2\sqrt { 2 } } =x+C } \)
Putting z = u² , we have
\(\int { \frac { dz }{ 4+2\sqrt { 2 } } =\frac { udu }{ u+2 } =u-2|u+2|+C } \)
or \(\sqrt z\) - 2 In(\(\sqrt z\) + 2) = x+C
from which on substituting z = 4x + 2y −1, we have the general solution
\(\sqrt { 4x+2y-1 } -2\quad In(\sqrt { 4x+2y-1 } +2)=x+C\)

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)