Ex 9.4 Class 12 Maths Ncert Solutions

\(\Rightarrow \frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}\)
\(\Rightarrow \frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\)
\(\Rightarrow \frac{d y}{d x}=\tan ^{2} \frac{x}{2}\)
\(\Rightarrow \frac{d y}{d x}=\left(\sec ^{2} \frac{x}{2}-1\right)\)
\(\Rightarrow d y=\left(\sec ^{2} \frac{x}{2}-1\right) d x\)
\(\Rightarrow \int d y=\int \sec ^{2} \frac{x}{2} d x-\int d x\)
\(\Rightarrow y=2 \tan ^{1} \frac{x}{2}-x+c\)

\(\Rightarrow \frac{d y}{d x}=\sqrt{4-y^{2}}\)
\(\Rightarrow \frac{d y}{\sqrt{4-y^{2}}}=d x\)
\(\Rightarrow \int \frac{d y}{\sqrt{4-y^{2}}}=\int d x\)
\(\Rightarrow \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}\)
\(\Rightarrow \sin ^{-1} \frac{y}{2}=x+c\)

\(\Rightarrow \frac{d y}{d x}+y=1\)
\(\Rightarrow d y=(1-y) d x\)
Separating variables
\(\Rightarrow \frac{d y}{1-y}=d x\)
\(\Rightarrow \int \frac{d y}{1-y}=\int d x\)
\(\Rightarrow-\log (1-{y})=x+\log {c}\)
\(\Rightarrow-\log (1-{y})-\log {c}=x\)
\(\Rightarrow \log (1-{y}) {c}=-x\)
\(\Rightarrow(1-y) c=e^{-x}\)
Or
\(\Rightarrow(1-y)=\frac{1}{c} e^{-x}\)
\(\Rightarrow y=1+\frac{1}{c} e^{-x}\)

\( \sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y \)
Dividing both sides by \( (\tan x)(\tan {y}) \)
\( \therefore \frac{\sec ^{2} x \tan y d x}{\tan x \tan y}+\frac{\sec ^{2} y \tan x d y}{\tan x \tan y}=0 \)
\( \Rightarrow \frac{\sec ^{2} x d x}{\tan x}+\frac{\sec ^{2} y d y}{\tan y}=0 \)
Integrating both sides,
\( \Rightarrow \int \frac{\sec ^{2} x d x}{\tan x}=\int \frac{\sec ^{2} y d y}{\tan y} \)
\( \Rightarrow \) let \( \tan x={t} \ \& \ \tan {y}={u} \)
\( \therefore \sec ^{2} x d x=d t \ \& \ \sec ^{2} y d y=d u \)
\( \therefore \int \frac{d t}{t}=-\int \frac{d u}{u n} \)
\( \Rightarrow \log {t}=-\log {u}+\log {c} \)
Or,
\( \Rightarrow \log (\tan x)=-\log (\tan {y})+\log {c} \)
\( \Rightarrow \log \tan x=\log \frac{c}{\tan y} \)
Or
\(\Rightarrow(\tan x)(\tan {y})={c}\)

\(\Rightarrow\left({e}^{x}+{e}^{-x}\right) {dy}-\left({e}^{x}-{e}^{-x}\right) {dx}=0\)
\(\Rightarrow d y=\frac{\left(e^{x}-e^{-x}\right) d x}{e^{x}+e^{-x}}\)
Let \( \left(e^{x}+e^{-x}\right)={t} \)
Then, \( \left(e^{x}-e^{-x}\right) d x=d t \)
\(\therefore y=\int \frac{d t}{t}\)
\(\because \int \frac{d x}{x}=\log x\)
So,
\(\Rightarrow {y}=\log {t}\)
Or,
\(\Rightarrow y=\log \left(e^{x}-e^{-x}\right)+c\)

\(\Rightarrow \frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)\)
Separating variables,
\(\Rightarrow \frac{d y}{1+y^{2}}=d x\left(1+x^{2}\right)\)
Integrating both sides,
\(\Rightarrow \int \frac{d y}{1+y^{2}}=\int d x+\int x^{2} d x\)
\(\Rightarrow \tan ^{-1} y=x+\frac{x^{3}}{3}+c\)

\(y \log y d x-x d y=0\)
\(\Rightarrow(y \log y) d x=x d y\)
Separating variables,
\(\Rightarrow \frac{d x}{x}=\frac{d y}{y \log y}\)
\(\Rightarrow \operatorname{let} \log {y}={t}\)
\(\Rightarrow \frac{1}{y} d y=d t\)
\(\Rightarrow \log x=\int \frac{d t}{t}\)
\(\Rightarrow \log x+\log {c}=\log {t}\)
\(\text {Or, }\)
\(\Rightarrow \log x+\log {c}=\log (\log {y})\)
\(\Rightarrow \log {cx}=\log {y}\)
\(\text { Or, }\)
\(\Rightarrow \log {y}={cx}\)
\(\Rightarrow y=e^{c x}\)

\(\Rightarrow x^{5} \frac{d y}{d x}=-y^{5}\)
Separating variables,
\(\Rightarrow \frac{d y}{y^{5}}=\frac{-d x}{x^{5}}\)
Or,
\(\Rightarrow \frac{d y}{y^{5}}=\frac{-d x}{x^{5}}=0\)
Let "\( a \)" be a constant,
\(\Rightarrow \int y^{-5} d y+\int x^{-5} d x=a\)
\(\Rightarrow-4 y^{-4}-4 x y^{-4}+c=a\)
Or,
\(\Rightarrow-x^{-4}-y^{-4}=c\)
Or,
\(\Rightarrow \frac{1}{x^{4}}+\frac{1}{x^{4}}=c\)

\(\Rightarrow \frac{d y}{d x}=\sin ^{-1} x\)
Separating variables,
\(\Rightarrow d y=\sin ^{-1} x d x\)
Integrating both sides,
\(\Rightarrow \int d y=\int \sin ^{-1} x d x\)
Now to integrate \( \sin ^{-1} x \) we have to multiply it by 1 because,
\(\because\left\{\int u . v d x=u \int v d x-\int\left(\frac{d}{d x} u\right)\left(\int v d x\right) d x\right\} \quad \ldots\ldots\text{(i)}\)
So,
\(\Rightarrow y=\int 1 \cdot \sin ^{-1} x d x\)
According to (i)
Let \(u\) be \( \sin ^{-1} x \) and \(v\) be \(1\)
We can take the values of \(u\) and \(v\) from the formula (I.L.A.T.E)
\(\therefore {y}=\left\{\sin ^{-1} x \int 1 . {dx}-\int\left(\frac{{d}}{{dx}} \sin ^{-1} x\right)\left(\int 1 . {dx}\right) {dx}\right\}\)
So,
\(\Rightarrow y=x \sin ^{-1} x-\int \frac{x}{\sqrt{1-x^{2}}} d x\)
\(\Rightarrow \text {let } 1-x^{2}={t}\)
\(\Rightarrow-2 x {dx}={dt}\)
\(\Rightarrow x {dx}=-\frac{d t}{2}\)
\(\Rightarrow y=x \sin ^{-1} x+\int \frac{1}{2 \sqrt{t}} d t\)
Or,
\(\Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \int t^{-\frac{1}{2}} d t\)
\(\Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \sqrt{t}+c\)
putting the value of \( t \),
\(\Rightarrow y=x^{-1} x+\sqrt{1+x^{2}}+c\)

\(e^{x} \tan y \ d x+1\left(1-e^{x}\right) \sec ^{2} y \ d y=0\)
\(\Rightarrow\left(1-e^{x}\right) \sec ^{2} y \ d y=-e^{x} \tan y \ d y=0\)
Separating the variables,
\(\Rightarrow \frac{\sec ^{2} y}{\tan y} d y=-\frac{e^{x}}{1-e^{x}} d x\)
Let \( \tan y=t \& 1-e^{x}=u \)
Then,
\(\left(\sec ^{2} y \ d y=d t\right) \&\left(e^{x} {dx}={du}\right)\)
Then,
\(\therefore \int \frac{d t}{t}=\int \frac{d u}{u}\)
Or,
\(\Rightarrow \log {t}=\log {u}+\log {c}\)
Substituting the values of \( t \) and \( u \) on above equation.
\(\Rightarrow \log (\tan y)=\log \left(1-e^{x}\right)+\log c\)
\(\Rightarrow \log \tan y=\log c\left(1-e^{x}\right)\)
Or,
\(\Rightarrow \tan y=c\left(1-e^{x}\right)\)

\(\Rightarrow\left(x^{3}+x^{2}+x+1\right) \frac{d y}{d x}=2 x^{2}+x\)
Separating variables,
\(\Rightarrow d y=\frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)} d x\)
Integrating both sides,
\(\Rightarrow \int d y=\int \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)} d x\quad \ldots\ldots\text{(i)}\)
Integrating it partially,
\(\Rightarrow \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1}\)
\(\Rightarrow \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{A x^{2}+A(B x+C)(x+1)}{(x+1)\left(x^{2}+1\right)}\)
\(\Rightarrow 2 x^{2}+x=A x^{2}+A+B x+C x+C\)
\(\Rightarrow 2 x^{2}+x=(A+B) x^{2}+(B+C) x+A+C\)
Now comparing the coefficients of
\(\Rightarrow A+B=2\)
\(\Rightarrow B+C=1\)
\(\Rightarrow A+C=0\)
Solving them we will get the values of \( {A}, {B}, {C} \)
\({A}=\frac{1}{2}, {~B}=\frac{3}{2}, {C}=-\frac{1}{2}\)
Putting the values of \( {A}, {B}, {C} \) in \(\quad \ldots\ldots\text{(i)}\)
\(\Rightarrow \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{1}{2} \frac{1}{(x+1)}+\frac{1}{2} \frac{3 x-1}{x^{2}+1} d x\)
So,
\(\Rightarrow \int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^{2}+1} d x\)
\(\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{2} \int \frac{x}{x^{2}+1} d x-\frac{1}{2} \int \frac{d x}{x^{2}+1}\)
\(\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \int \frac{2 x}{x^{2}+1} d x-\frac{1}{2} \tan ^{-1} x\)
Let \( x^{2}+1=t \)
Then, \( 2 {xdx}={dt} \)
\(\therefore \frac{3}{4} \int \frac{2 x}{x^{2}+1} d x=\frac{3}{4} \int \frac{d t}{t}\)
So, \( {I}=\frac{3}{4} \log t \)
Or,
\({I}=\frac{3}{4} \log \left(x^{2}+1\right)\)
\(\Rightarrow y=\frac{1}{2} \log (x+1)+\frac{3}{4} \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+c\)
Or,
\(\Rightarrow y=\frac{1}{4}\left[2 \log (x+1)+3 \log \left(x^{2}+1\right)-\frac{1}{2} \tan ^{-1} x+c\right]\)
\(\Rightarrow y=\frac{1}{4}\left[\log \left\{(x+1)^{2}\left(x^{2}+1\right)\right\}\right]-\frac{1}{2} \tan ^{-1} x+c\quad \ldots\ldots\text{(ii)}\)
Now, we are given that \( {y}=1 \) when \( x=0 \)
\(\therefore 1=\frac{1}{4}\left[\log \left\{(0+1)^{2}\left(0^{2}+1\right)\right\}\right]-\frac{1}{2} \tan ^{-1} 0+c\)
\({I}=\frac{1}{4} \times 0-\frac{1}{2} \times 0+c\)
So,
\({C}=1\)
Putting the value of \(c\) in (ii)
\(y=\frac{1}{4}\left[\log \left\{(x+1)^{2}\left(x^{2}+1\right)\right\}\right]-\frac{1}{2} \tan ^{-1} x+1\)

\(x\left(x^{2}-1\right) \frac{d y}{d x}=1\)
Separating variables,
\(\Rightarrow d y=\frac{d x}{x\left(x^{2}+1\right)}\)
Or,
\(\Rightarrow d y=\frac{d x}{x(x+1)(x-1)}\)
Integrating both sides,
\(\Rightarrow \int d y=\int \frac{d x}{x(x+1)(x-1)}\quad \ldots\ldots\text{(i)}\)
Now let,
\(\Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}\quad \ldots\ldots\text{(ii)}\)
\(\Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A(x-1)(x+1)+B(x)(x-1)+C(x)(x+1)}{x(x+1)(x-1)}\)
Or,
\(\Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{(A+B+C) x^{2}+(B-C) x-A}{x(x+1)(x-1)}\)
Now comparing the values of \( \mathrm{A}, \mathrm{B}, \mathrm{C} \)
\( A+B+C=0 \)
\( B-C=0 \)
\( \mathrm{A}=-1 \)
Solving these we will get that \( B=\frac{1}{2} \) and \( C=\frac{1}{2} \)
Now putting the values of \( A, B, C \) in \(\quad \ldots\ldots\text{(ii)}\)
\(\Rightarrow \frac{1}{x(x+1)(x-1)}=-\frac{1}{x}+\frac{1}{2}\left(\frac{1}{x+1}\right)+\frac{1}{2}\left(\frac{1}{x-1}\right)\)
Now integrating it,
\(\Rightarrow \int d y=-\int \frac{1}{x} d x+\frac{1}{2} \int\left(\frac{1}{x+1}\right) d x+\frac{1}{2} \int\left(\frac{1}{x-1}\right) d x\)
\(\Rightarrow y=-\log x+\frac{1}{2} \log (x+1)+\frac{1}{2} \log (x-1)+\log c\)
\(\Rightarrow y=\frac{1}{2} \log \left[\frac{c^{2}(x-1)(x+1)}{x^{2}}\right]\quad \ldots\ldots\text{(iii)}\)
Now we are given that \( y=0 \) when \( x=2 \)
\(0=\frac{1}{2} \log \left[\frac{c^{2}(2-1)(2+1)}{4}\right]\)
\(\Rightarrow \log \frac{3 c^{2}}{4}=0\)
Or,
\(\Rightarrow \frac{3 c^{2}}{4}=1\)
\(\Rightarrow 3 c^{2}=4\)
\(\Rightarrow c^{2}=\frac{ 4 }{ 3 }\)
Now putting the value of \( c^{2} \) in \(\quad \ldots\ldots\text{(iii)}\)
Then,
\(\Rightarrow y=\frac{1}{2} \log \left[\frac{4(x-1)(x+1)}{3 x^{2}}\right]\)
\(\Rightarrow y=\frac{1}{2} \log \left[\frac{4\left(x^{2}-1\right)}{3 x^{2}}\right]\)

\(\Rightarrow \cos \frac{d y}{d x}=a\)
\(\Rightarrow \frac{d y}{d x}=\cos ^{-1} a\)
Separating variables,
\(\Rightarrow d y=\cos ^{-1} a d x\)
Integrating both sides,
\(\Rightarrow \int d y=\cos ^{-1} a \int d x\)
\(\Rightarrow y=x \cos ^{-1} a+c\quad \ldots\ldots\text{(i)}\)
Now \( y=2 \) when \( x=0 \)
\(\Rightarrow 2=0+\mathrm{c}\)
\(\Rightarrow c=2\)
Putting the value of \(\quad \ldots\ldots\text{(ii)}\)
\(\Rightarrow y=x \cos ^{-1} a+2\)

\(\frac{d y}{d x}=y \tan x\)
Separating variables,
\(\Rightarrow \frac{d y}{y}=\tan x d x\)
Integrating both sides,
\(\Rightarrow \int \frac{d y}{y}=\int \tan x d x\)
\(\Rightarrow \log {y}=-\log (\cos x)+\log {c}\)
Or,
\(\Rightarrow \log {y}=\log (\sec x)+\log {c}\)
\(\Rightarrow \log {y}=\log {c}(\sec x)\)
\(\Rightarrow {y}={c}(\sec x)-{i})\)
Now we are given that \( {y}=1 \) when \( x=0 \)
\(\Rightarrow 1={c}(\sec 0)\)
\(\Rightarrow 1={c} \times 1\)
\(\Rightarrow {c}=1\)
Putting the value of \( c \) in (i)
\(\Rightarrow y=\sec x\)

To find the equation of a curve that passes through point \( (0,0) \) and has differential equation \( y^{\prime}=e^{x} \sin x \).
So, we need to find the general solution of the given differential equation and the put the given point in to find the value of constant.
So, \( \Rightarrow \frac{d y}{d x}={e}^{x} \sin x \).
Separating variables,
\(\Rightarrow d y={e}^{x} \sin {xdx}\)
Integrating both sides,
\(\Rightarrow \int d y=\int e^{x} \sin x d x\quad \ldots\ldots\text{(i)}\)
\( \left\{ \text{Using the formula, }\int u \cdot v d x=u \int v d x-\int\left\{\frac{d}{d x} u \int v d x\right\} d x\right\} \)
Now let \( I=\int e^{x} \sin x d x \)
\(\Rightarrow {I}=\sin x \int e^{x} d x-\int\left(\frac{d}{d x} \sin x \iint e^{x} d x\right) d x\)
\(\Rightarrow {I}=e^{x} \sin x-\int \cos x e^{x} d x\)
Now integrating \( \int e^{x} \cos x d x \) same way as integrating for \( \int e^{x} \sin x d x \)
\(\Rightarrow {I}=e^{x} \sin x d x-\left[\cos x \int e^{x} d x+\int \sin x e^{x} d x\right]\)
\( \Rightarrow {I}=e^{x} \sin x-e^{x} \cos x-{I} \quad \) (from i)
Or,
\(\Rightarrow 2 {I}=e^{x} \sin x-e^{x} \cos x\)
\(\Rightarrow 2 {I}=e^{x}(\sin x-\cos x)\)
\(\Rightarrow {I}=e^{x} \frac{(\sin x-\cos x)}{2}\)
Or,
\(\Rightarrow y=e^{x} \frac{(\sin x-\cos x)}{2}+c\quad \ldots\ldots\text{(ii)}\)
Now we are given that the curve passes through point \( (0,0) \)
\(\therefore 0=e^{0} \frac{(\sin 0-\cos 0)}{2}+c\)
\(\Rightarrow 0=\frac{1(0-1)}{2}+c\)
\(\Rightarrow c=\frac{1}{2}\)
Putting the value of C in (ii)
So,
\(\Rightarrow y=e^{x} \frac{(\sin x-\cos x)}{2}+\frac{1}{2}\)
\(\Rightarrow 2 y=e^{x}(\sin x-\cos x)+1\)
So the required equations become,
\(\Rightarrow 2 y-1=e^{x}(\sin x-\cos x)\)

For this question, we need to find the particular solution at point \( (1,-1) \) for the given differential equation.
Given differential equation is
\(\Rightarrow x y \frac{d y}{d x}=(x+2)(y+2)\)
Separating variables,
\(\Rightarrow \frac{y}{y+2} d y=\frac{(x+2) d x}{x}\)
Or,
\(\Rightarrow\left(1-\frac{2}{y+2}\right) d y=\left(1+\frac{2}{x}\right) d x\)
Integrating both sides,
\(\Rightarrow \int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x\)
\(\Rightarrow \int d y-2 \int \frac{1}{y+2} d y=\int d x+2 \int \frac{1}{x} d x\)
\(\Rightarrow y-2 \log (y+2)=x+2 \log x+c\)
Now separating like terms on each side,
\(\Rightarrow y-x-c=2 \log x+2 \log (y+2)\)
\(\Rightarrow y-x-c=\log x^{2}+\log (y+2)^{2}\)
Or,
\(\Rightarrow y-x-c=\log \left\{x^{2}(y+2)^{2}\right\}\quad \ldots\ldots\text{(i)}\)
Now we are given that, the curve passes through \( (1,-1) \)
\( \therefore \) putting the values of \( x \& \mathrm{y} \), to find the value of \(c \).
\(\Rightarrow-1-1-c=\log \left\{1(-1+2)^{2}\right\}\)
\(\Rightarrow-2-c=\log (1)\)
\(\Rightarrow \mathrm{c}=-2+0\quad (\because \log (1)=0)\)
So \( \mathrm{c}=-2 \)
Putting the value of \(c\) in
\(y-x-c=\log \left\{x^{2}(y+2)^{2}\right\}\)
\(y-x-2=\log \left\{x^{2}(y+2)^{2}\right\}\)

We know that slope of a tangent is \( =\frac{d y}{d x} \).
So we are given that the product of the slope of its tangent and \( y \) coordinate of the point is equal to the \(x\) coordinate of the point.
\(\therefore y \frac{d y}{d x}=x\)
Now separating variables,
\(\Rightarrow y d y=x d x\)
Integrating both sides,
\(\Rightarrow \int y d y=\int x d x\)
\(\Rightarrow \frac{y^{2}}{2}=\frac{x^{2}}{2}+c\)
\(\Rightarrow y^{2}-x^{2}=2 c\quad \ldots\ldots\text{(i)}\)
Now the curve passes through \( (0,-2) \).
\(\therefore 4-0=2 c\)
\(\Rightarrow \mathrm{c}=2\)
Putting the value of \(c\) in \(\quad \ldots\ldots\text{(i)}\)
\(\Rightarrow y^{2}-x^{2}=4\)

We know that \( (x, {y}) \) is the point of contact of curve and its tangent.
slope \((m_{1})\) for line joining \( (x, y) \) and \( (-4,-3) \) is \( \frac{y+3}{x+4}\quad \ldots\ldots\text{(i)} \)
Also we know that slope of tangent of a curve is \( \frac{d y}{d x} \).
\( \therefore \) slope \( \left({m}_{2}\right) \) of tangent \( =\frac{d y}{d x} \quad \ldots\ldots\text{(ii)}\)
Now, according to the question,
\(({m}_{2})=2({m}_{1})\)
\(\Rightarrow \frac{d y}{d x}=\frac{2(y+3)}{x+4}\)
Separating variables,
\(\Rightarrow \frac{d y}{y+3}=\frac{2 d x}{x+4}\)
Integrating both sides,
\(\Rightarrow \int \frac{d y}{y+3}=2 \int \frac{d x}{x+4}\)
\(\Rightarrow \log (y+3)=2 \log (x+4)+\log {c}\)
\(\Rightarrow \log ({y}+3)=2 \log (x+4)^{2}\)
\(\Rightarrow y+3=c(x+4)^{2}\quad \ldots\ldots\text{(iii)}\)
Now, this equation passes thorough the point \( (-2,1) \).
\(\Rightarrow 1+3=c(-2+4)^{2}\)
\(\Rightarrow 4=4 c\)
Substitute the value of \(c\) in (iii)
\(\Rightarrow y+3=(x+4)^{2}\)

Let the rate of change of the volume of the balloon be \(k \). (\( k\) is a constant)
\(\therefore \frac{d y}{d t}=k\)
Or,
\(\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=k\left\{\text {volume of sphere }=\frac{4}{3} \pi r^{3}\right\}\)
\(\Rightarrow \frac{4}{3} \pi 3 r^{2} \frac{d r}{d t}=k\)
\(\Rightarrow 4 \pi r^{2} d r=k d t\)
Integrating both sides,
\(\Rightarrow 4 \pi \int r^{2} d r=k \int d t\)
\(\Rightarrow \frac{4 \pi r^{3}}{3}=k t+c\quad \ldots\ldots\text{(i)}\)
Now, given that
At \( {t}=0, {r}=3 \) :
\(\Rightarrow 4 \pi \times 33=3({k} \times 0+{c})\)
\(\Rightarrow 108 \pi=3 {c}\)
\(\Rightarrow {c}=36 \pi\)
At \( {t}=3, {r}=6 \) :
\(\Rightarrow 4 \pi \times 6^{3}=3(k \times 3+c)\)
\(\Rightarrow {k}=84 \pi\)
Substituting the values of \( k \) and \(c\) in (i)
\(\Rightarrow 4 \pi r^{3}=3(84 \pi t+36 \pi)\)
\(\Rightarrow 4 \pi r^{3}=4 \pi(63 t+27)\)
\(\Rightarrow r^{3}=63 t+27\)
\(\Rightarrow r^{3}=\sqrt[3]{63 t+27}\)
So the radius of balloon after \( t \) seconds is \( \sqrt[3]{63 t+27} \)

let \(t\) be time
\(p\) be principal
\( r \) be rate of interest
according the information principal increases at the rate of \( {r} \% \) per year.
\(\therefore \frac{d p}{d t}=\left(\frac{r}{100}\right) p\)
Separating variables,
\(\Rightarrow \frac{d p}{p}=\left(\frac{r}{100}\right) d t\)
Integrating both sides,
\(\Rightarrow \int \frac{d p}{p}=\frac{r}{100} \int d t\)
\(\Rightarrow \log p=\frac{r t}{100}+k\)
\(\Rightarrow p=e^{\frac{r t}{100}+k}\quad \ldots\ldots\text{(i)}\)
Given that \( {t}=0, {p}=100 \).
\(\Rightarrow 100=e^{k}\quad \ldots\ldots\text{(ii)}\)
Now, if \( {t}=10 \), then \( {p}=2 \times 100=200 \)
So,
\(\Rightarrow 200=e^{\frac{r t}{10}+k}\)
\(\Rightarrow 200=e^{\frac{r t}{10}} \cdot e^{k}\)
\(\Rightarrow 200=e^{\frac{r t}{10}} \times 100 \text { from (ii) }\)
\(\Rightarrow e^{\frac{r}{10}}=2\)
\(\Rightarrow \frac{r}{10}=\log 2\)
\(\Rightarrow r=6.93\)
So \( r \) is \( 6.93 \% \).

Let \(p\) and \(t\) be principal and time respectively.
Given that principal increases continuously at rate of \( 5 \% \) per year.
\(\therefore \frac{d p}{d t}=\left(\frac{5}{100}\right) p\)
Separating variables,
\(\Rightarrow \frac{d p}{p}=\left(\frac{p}{100}\right)\)
Integrating both sides,
\(\Rightarrow \int \frac{d p}{p}=\frac{1}{20} \int d t\)
\(\Rightarrow \log p=e^{\frac{t}{20}+c}\quad \ldots\ldots\text{(i)}\)
When \( {t}=0, {p}=1000 \)
\(\Rightarrow 1000=e^{c}\)
At \( {t}=10 \)
\(\Rightarrow p=e^{\frac{t}{20}+c}\)
\(\Rightarrow p=e^{0.5} \times e^{c}\)
\(\Rightarrow p=1.648 \times 1000\left(e^{0.5}=1.648\right)\)
\(\Rightarrow p=1648\)
So after 10 years the total amount would be Rs. 1648

let \( y \) be the number of bacteria at any instant \( t \).
Given that the rate of growth of bacteria is proportional to the number present
\(\therefore \frac{d y}{d t} \propto y\)
\(\Rightarrow \frac{d y}{d t}=k y \ ({k} \text { is a constant})\)
Separating variables,
\(\Rightarrow \frac{d y}{d t}=k d t\)
Integrating both sides,
\(\Rightarrow \int \frac{d y}{y}=k \int d t\)
\(\Rightarrow \log {y}={kt}+{c}\quad \ldots\ldots\text{(i)}\)
Let \( y^{\prime} \) be the number of bacteria at \( {t}=0 \).
\(\Rightarrow \log {y}^{\prime}={c}\)
Substituting the value of \(c\) in (i)
\(\Rightarrow \log {y}={kt}+\log {y}^{\prime}\)
\(\Rightarrow \log {y}-\log {y}^{\prime}={kt}\)
\(\Rightarrow \log \frac{y}{y^{\prime}}=k t\quad \ldots\ldots\text{(ii)}\)
Also, given that number of bacteria increases by \( 10 \% \) in 2 hours.
\(\Rightarrow y=\frac{110}{100} y^{\prime}\)
\(\Rightarrow \frac{y}{y^{\prime}}=\frac{11}{10}\quad \ldots\ldots\text{(iii)}\)
Substituting this value in (ii)
\( \Rightarrow k \times 2=\log \frac{11}{10} \)
\( \Rightarrow k=\frac{1}{2} \log \frac{11}{10} \)
So, (ii) becomes
\( \Rightarrow \frac{1}{2} \log \frac{11}{10} \times t=\log \frac{y}{y^{\prime}} \)
\( \Rightarrow t=\frac{2 \log \frac{y}{y^{\prime}}}{\log \frac{11}{10}} \quad \ldots\ldots\text{(iv)}\)
Now, let the time when number of bacteria increase from 100000 to 200000 be \( t^{\prime} \).
\( \Rightarrow y=2 y^{\prime} \) at \( {t}={t}^{\prime} \)
So from (iv)
\( \Rightarrow t^{\prime}=\frac{2 \log \frac{y}{y^{\prime}}}{\log \frac{11}{10}}=\frac{2 \log 2}{\log \frac{11}{10}} \)
So bacteria increases from 100000 to 200000 in \( \frac{2 \log 2}{\log \frac{11}{10}} \) hours.

\(\Rightarrow \frac{d y}{d x}=e^{x+y}\)
\(\Rightarrow \frac{d y}{d x}=e^{x} \times e^{y}\)
separating variables
\(\Rightarrow e^{-y} d y=e^{x} d x\)
Integrating both sides
\(\Rightarrow \int e^{-y} d y=\int e^{x} d x\)
\(\Rightarrow-e^{-y}=e^{x}+c\)
\(\Rightarrow e^{x}+e^{-y}=-c\)
Or,
\(e^{x}+e^{-y}=-c \ ({c} \text { is a constant})\)
So the correct option is A.