Let 'P' be the principal at any time t.
By the question.\(\frac { dP }{ dt } =\frac { 5 }{ 100 } P\)            
\(\Rightarrow \)\(\frac { dP }{ P } =\frac { 1 }{ 20 } dt\) | Variables Separable
Integrating, \(\int { \frac { dP }{ P } =\frac { 1 }{ 20 } \int { 1.dt+c } } \)
\(\Rightarrow \) \({ log }_{ e }P=\frac { 1 }{ 20 } t+c\)....(1)
\(\left[\therefore P>0\right]\)
When \(t=0,P=1000,{ log }_{ e }1000=0+c\Rightarrow c={ log }_{ e }1000.\)
Putting in (1), \({ log }_{ e }P=\frac { 1 }{ 20 } t+{ log }_{ e }1000\) ..(2)
When \(P=2000,{ log }_{ e }2000=\frac { 1 }{ 20 } t+{ log }_{ e }1000\)
\(\Rightarrow \)\(\frac { 1 }{ 20 } t={ log }_{ e }\frac { 2000 }{ 1000 } ={ log }_{ e }\quad 2\Rightarrow t=20{ log }_{ e }2.\)=
Hence, Rs 1000 double in 20 \({ log }_{ e }\) 2 years.

We know that the slope of the tangent to a curve is given by \(\frac{d y}{d x} \)
 so,\(\frac { dy }{ dx } =\frac { 2x }{ { y }^{ 2 } } \) ... (1)
Separating the variables, equation (1) can be written as
  \({ y }^{ 2 }dy=2xdx\) ... (2)
Integrating both sides of equation (2), we get
\(\int { { y }^{ 2 }dy=2\int { xdx+c } } \)
\(\Rightarrow \) \(\frac { { y }^{ 3 } }{ 3 } ={ x }^{ 2 }+c\) ...(3)
Substituting x = –2, y = 3 in equation (3), we get C = 5.
Substituting the value of C in equation (3), we get the equation of the required curve as         
\(\frac{y^3}{3}=x^2+5 \ \text { or } \ y=\left(3 x^2+15\right)^{\frac{1}{3}}\)

We have: \(\frac { dx }{ dy } =\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } \) .(1)
Here \(f\left( x,y \right) =\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } .\)
\(=\frac { 2x{ e }^{ \frac { x }{ y } }-y }{ 2y{ e }^{ \frac { x }{ y } } } ={ \lambda }^{ 0 }f\left( x,y \right) .\)
Thus \(f(x,y)\) is homogeneous function of degree 0.
To solve:
Put  \(\frac { x }{ y } =v\quad i.e\quad x=vy\)
So that   \(\frac { dx }{ dy } =v+y\frac { dy }{ dy } .\)     
\(\therefore\) (1) becomes : \(v+y\frac { dy }{ dx } =\frac { 2vy\quad { e }^{ v }-y }{ 2y\quad { e }^{ v } } \)
\(\Rightarrow \)  \(y\frac { dv }{ dy } =\frac { 2\quad v\quad { e }^{ v }-1 }{ 2\quad { e }^{ v } } -v\)
\(\Rightarrow\)  \(y\frac { dv }{ dy } =\frac { 2\quad v\quad { e }^{ v }-1 }{ 2\quad { e }^{ v } } -v\)
\(\Rightarrow \)  \(y\frac { dv }{ dy } =\frac { -1 }{ 2{ e }^{ v } } \)
\(\Rightarrow \)  \(2{ e }^{ v }dv=-\frac { dy }{ y } \)
Integrating,\(2\int { { e }^{ v }\ dv=-log|y|+c } \)
\(\Rightarrow \) \(2{ e }^{ v }=-log|y|+c\)
\(\Rightarrow \)\(2{ e }^{ \frac { x }{ y } }=-log|y|+c\) ...(2)
When \(x=0,y=1\)
\(\therefore\)  \(2\left( 1 \right) =-log|1|+c\)
\(\Rightarrow \) \(2=-0+c\Rightarrow =2.\)
Putting in (2),
\(2{ e }^{ \frac { x }{ y } }+log|y|=2,\)
Which is the required solution.

\(\Rightarrow \)  xy = \(\frac {{x}^{4}}{4} + c\)is the required solution.

Given differential equation is
\(\frac{d y}{d x}-\frac{y}{x}+\operatorname{cosec}\left(\frac{y}{x}\right)=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x}-\operatorname{cosec}\left(\frac{y}{x}\right)\) 
which is a homogeneous differential equation as \(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\) .
On putting \(y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}\) in equation (i), we get
\(v+x \frac{d v}{d x}=v-\operatorname{cosec} v\) 
\(\Rightarrow x \frac{d v}{d x}=-\operatorname{cosec} v \)
\(\Rightarrow \sin v d v=-\frac{d x}{x}\) 
On integrating both sides, we get
\(\int \sin v d v=-\int \frac{d x}{x}\) 
\(\Rightarrow -\cos v=-\log |x|+C\) 
\(\Rightarrow \cos v=\log |x|-C\) 
\(\Rightarrow \cos \left(\frac{y}{x}\right)=\log |x|-C \quad\left[\text { put } v=\frac{y}{x}\right]..(ii)\) 
Also, given y = 0, when x = 1
Then, \(\cos 0=\log 1-C \Rightarrow 1=0-C \Rightarrow C=-1\) 
So, equation (ii) becomes \(\cos \left(\frac{y}{x}\right)=\log |x|+1\) 
which is the required equation 

\(y=\frac{1}{5}(2 \sin x-\cos x)+\mathrm{C} e^{-2 x}\)

Given differential equation is
\(\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0  \)
\(\Rightarrow \frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}} \)
On separating the variables, we get
\(\frac{1}{\sqrt{1-y^{2}}} d y=-\frac{1}{\sqrt{1-x^{2}}} d x \)
On integrating both sides, we get
\(\int \frac{1}{\sqrt{1-y^{2}}} d y=-\int \frac{1}{\sqrt{1-x^{2}}} d x  \)
\(\Rightarrow \sin ^{-1} y=-\sin ^{-1} x+C \)
\(\Rightarrow \sin ^{-1} x+\sin ^{-1} y=C  \)
which is the required general solution.

\(\frac { dy }{ dx } =\sqrt { \frac { 1-\cos { x } }{ 1+\cos { x } } } \)
Integrating both the sides,
\(\int { dy } =\int { \sqrt { \frac { 1-\cos { x } }{ 1+\cos { x } } } } dx\)
\(\Rightarrow \int { dy } =\int { \sqrt { \frac { 2\sin ^{ 2 }{ \frac { x }{ 2 } } }{ 2\cos ^{ 2 }{ \frac { x }{ 2 } } } } } dx\)
\(\left\{ \because \quad \cos { x } =2\cos ^{ 2 }{ \frac { x }{ 2 } } -1\quad and\quad \cos { x } =1-2\sin ^{ 2 }{ \frac { x }{ 2 } } \right\} \)
\(\Rightarrow \int { dy } =\int { \tan { \frac { x }{ 2 } } } dx\)
\(\Rightarrow y=-2\log { \cos { \frac { x }{ 2 } } } +c\)
\(\Rightarrow \log { \cos { \frac { x }{ 2 } } } +c\)

y = mx + c
\(\Rightarrow \frac { dy }{ dx } =m\)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0\)

First of all we factorise RHS as \(\frac{dy}{dx}\)  = (1+x)(1+y) and then separate the variables.

Given differential equation is
\(\left(1+x^{2}\right) d y+2 x y d x=\cot x d x \quad[\because x \neq 0]\) 
Above equation can be rewritten as,
\(\left(1+x^{2}\right) d y+(2 x y-\cot x) d x=0\) 
\(\Rightarrow\left(1+x^{2}\right) d y=(\cot x-2 x y) d x\) 
On dividing both sides by \(1+x^{2}\) ,we get
\(d y=\frac{\cot x-2 x y}{1+x^{2}} d x\) 
\(\Rightarrow \frac{d y}{d x}=\frac{\cot x}{1+x^{2}}-\frac{2 x y}{1+x^{2}}\) 
\(\Rightarrow \frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{\cot x}{1+x^{2}}\) 
which is a linear differential equation of the form of
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=\frac{2 x}{1+x^{2}} \text { and } Q=\frac{\cot x}{1+x^{2}}\) 
Now,\(\mathrm{IF}=e^{\int P d x}=e^{\frac{1}{1+x^{2}} d x}=e^{\log \left|1+x^{2}\right|}=1+x^{2}\) 
\(\left[\because I_{1}=\int \frac{2 x}{1+x^{2}} d x,\right. \text { put } 1+x^{2}=t \Rightarrow 2 x d x=d t\) 
\(\left.\Rightarrow I_{1}=\int \frac{d t}{t}=\log |t|=\log \left|1+x^{2}\right|\right]\)
and the solution of linear differential equation is given by
\(y \times I F=\int(Q \times I F) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \frac{\cot x}{1+x^{2}} \times\left(1+x^{2}\right) d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\int \cot x d x+C\) 
\(\Rightarrow y\left(1+x^{2}\right)=\log |\sin x|+C\) 
\(\Rightarrow y=\frac{\log |\sin x|}{1+x^{2}}+\frac{C}{1+x^{2}}\) 
which is the required solution.

Given, \(\frac{d y}{d x}-(3 \cot x) y=\sin 2 x\) 
which is a linear differential equation of the form
\(\frac{d y}{d x}+P y=Q\) 
Here, \(P=-3 \cot x \text { and } Q=\sin 2 x\) 
Now, \(\mathrm{IF}=e^{\int P d x}=e^{-3 \int \cot x d x}=e^{-3 \log (\sin x)}=e^{\log (\sin x)^{-2}}\) 
\(=\frac{1}{\sin ^{3} x}\) 
and the required solution is given by
\(\boldsymbol{y} \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C\) 
\(\Rightarrow y \times \frac{1}{\sin ^{3} x}=\int \frac{1}{\sin ^{3} x} \sin 2 x d x+C\) 
\(\Rightarrow y \times \frac{1}{\sin ^{3} x}=2 \int \frac{\sin x \cos x}{\sin ^{3} x} d x+C\) 
\([\because \sin 2 x=2 \sin x \cos x]\) 
\(\Rightarrow \frac{1}{\sin ^{3} x} \times y=2 \int \frac{\cos x}{\sin ^{2} x} d x+C\) 
\(\Rightarrow \frac{y}{\sin ^{3} x}=-2 \operatorname{cosec} x+C\) 
\(\Rightarrow y=-2\left(\frac{1}{\sin x} \times \sin ^{3} x\right)+C \sin ^{3} x\) 
\(\Rightarrow y=-2 \sin ^{2} x+C \sin ^{3} x\) 
Also, given \(y=2 \text { and } x=\frac{\pi}{2}\), therefore from Eq. (i), we get
\(2=-2 \sin ^{2}\left(\frac{\pi}{2}\right)+C \sin ^{3}\left(\frac{\pi}{2}\right)\) 
\(\Rightarrow 2=-2+C \)
\(\Rightarrow C=4 \) 
On putting the value of C in Eq. (i), we get
\(y=-2 \sin ^{2} x+4 \sin ^{3} x \Rightarrow y=4 \sin ^{3} x-2 \sin ^{2} x\) 
which is the required solution.

 (i) (c) : The given differential equation can be written as \( \frac{d y}{d x}+2 y \cot x=\operatorname{cosec} x\)
\(\therefore \text { I.F. }=e^{\int 2 \cot x d x}=e^{2 \log |\sin x|}=(\sin x)^{2} \)
\(\therefore \quad \lambda=2\)
(ii) (c) : We have, \(\left(1-x^{2}\right) \frac{d y}{d x}-x y=1\)
\(\Rightarrow \frac{d y}{d x}-\frac{x}{1-x^{2}} \cdot y=\frac{1}{1-x^{2}} \)
\(\therefore \text { I.F. }=e^{-\int \frac{x}{1-x^{2}} d x}=e^{\frac{1}{2} \int \frac{-2 x}{1-x^{2}} d x}\)
\(=e^{\frac{1}{2} \log \left(1-x^{2}\right)}=e^{\log \left(1-x^{2}\right)^{\frac{1}{2}}}=\sqrt{1-x^{2}}\)
(iii) (b) : We have, \(\frac{d y}{d x}+y=e^{-x} \)
 It is a linear differential equation with I.F. = \(e^{\int d x}=e^{x}\)
Now, solution is \(y \cdot e^{x}=\int e^{x} \cdot e^{-x} d x+c\)
\(\Rightarrow y e^{x}=\int d x+c \Rightarrow y e^{x}=x+c \Rightarrow y=x e^{-x}+c e^{-x}\)
\(\because y(0)=0 \Rightarrow c=0 \quad \therefore y=x e^{-x}\)
(iv) (a) : We have,\( \frac{d y}{d x}+y \tan x=\sec x\)
It is a linear differential equation with I.F. = \(e^{\int \tan x d x}=e^{\log |\sec x|}=\sec x\)
Now, solution is \(y \sec x=\int \sec ^{2} x d x+c\)
\(\Rightarrow y \sec x=\tan x+c\)
(v) (c) : We have, \(\frac{d y}{d x}-3 y=\sin 2 x\)
It is a linear differential equation with \(\text { I.F. }=e^{\int-3 d x}=e^{-3 x}\)

(a) Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.