Differential Equations

The given D.E. is

$\begin{array}{l}\frac{dy}{dx}=e^{x+y}\\ \Rightarrow \frac{dy}{dx}=e^x.e^y\\ \Rightarrow \frac{dy}{e^y}=e^{x}dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{e^y}={\displaystyle \int e^{x}dx}}\\ \Rightarrow \frac{e^{-y}}{-1}=e^x+c_1\\ \Rightarrow e^{-y}=-e^x-c_1\\ \Rightarrow e^{-y}+e^x=c,where,c=-c_1\end{array}$

$\therefore$ Option (A) is correct.

Let 'x' be the number of bacteria present in instantaneous time t.

Then, $\frac{dx}{dt}\propto x$

$\Rightarrow \frac{dx}{dt}=kx,where,k=$ constant of proportionality.

$\Rightarrow \frac{dx}{x}=kdt$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dx}{x}={\displaystyle \int kdt}}\\ \Rightarrow logx=kt+c\end{array}$

Given, at $t=0,x=x_0(say)then,$

$log{x}_0=c\left(Initial,x_0=100000\right)$

So, the differential equation is

$\begin{array}{l}logx=kt+log{x}_0\\ \Rightarrow logx-log{x}_0=kt\\ \Rightarrow log\frac{x}{x_0}=kt\end{array}$

As the bacteria number increased by 10% in 2 hours.

The number of bacteria increased in 2hours $=10\mathrm{\%}*100000=10000$

Hence, at t=2,

$x=100000+10000=110000$

So, $\begin{array}{l}log\frac{110000}{100000}=2k\\ \Rightarrow k=\frac{1}{2}log\left(\frac{11}{10}\right)\end{array}$

Hence, $log\frac{x}{x_0}=\left[\frac{1}{2}log\frac{11}{10}\right]*t$

$when,x=200000,$ then we get,

$log\frac{200000}{100000}=\frac{1}{2}log\frac{11}{10}*t$

$\begin{array}{l}\Rightarrow 2log2=log\left(\frac{11}{10}\right)*t\\ \Rightarrow t=\frac{2log2}{log\left(\frac{11}{10}\right)}hours\end{array}$

Let P and t the principal and time respectively.

Then, increase in principal $\frac{dP}{dt}=P*5\mathrm{\%}$

$\Rightarrow \frac{dP}{P}=\frac{5}{100}dt$

Integrating both sides,

$\begin{array}{l}\Rightarrow {\displaystyle \int \frac{dP}{P}={\displaystyle \int \frac{1}{20}dt}}\\ \Rightarrow logP=\frac{t}{20}+c\\ \Rightarrow P=e^{\frac{t}{20}+c}\end{array}$

At, t=0, P=1000

So,  $\begin{array}{l}1000=e^{\frac{0}{20}+c}\\ \Rightarrow e^c=1000\end{array}$

And at t=10,

$\begin{array}{l}P=e^{\frac{10}{100}+c}=e^{0.5}.e^c\\ \Rightarrow P=1.648*1000=1648\end{array}$

P = ?1648

Let P, r and t be the principal rate and time respectively.

Then, increase in principal $\frac{dP}{dt}=P*r\mathrm{\%}$

$\begin{array}{l}\Rightarrow \frac{dP}{dt}=P.\frac{r}{100}\\ \Rightarrow \frac{dP}{P}=\frac{r}{100}dt\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dP}{P}={\displaystyle \int \frac{r}{100}dt}}\\ \Rightarrow logP=\frac{rt}{100}+c\\ \Rightarrow P=e^{\frac{rt}{100}+c}\end{array}$

Given at t=0,P=100

So, $100=e^{\frac{r*0}{100}+c}$

$\begin{array}{l}\Rightarrow 100=e^0*e^c\\ \Rightarrow e^c=100\left(?e^0=1\right)\end{array}$

And at $t=10,P=2*100=200$

So, $200=e^{\frac{r10}{100}+c}$

$\begin{array}{l}\Rightarrow 200=e^{\frac{r}{10}}.e^e\\ \Rightarrow e^{\frac{r}{10}}=\frac{200}{100}=2\end{array}$

$\begin{array}{l}\Rightarrow \frac{r}{10}=log2\\ \Rightarrow \frac{r}{10}=0.6931\\ \Rightarrow r=6.931\end{array}$

Hence, the rate is 6.931%

Let 'r' and U be the radius and volume of the spherical balloon.

Then, $\frac{dU}{dt}=k,$ k = constant

$\begin{array}{l}\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=k\\ \Rightarrow 4\pi r^2\frac{dr}{dt}=k\\ \Rightarrow 4\pi r^{2}dr=kdt\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int 4\pi r^{2}dr={\displaystyle \int kdt}}\\ \Rightarrow \frac{4}{3}\pi r^3=kt+c\end{array}$

Given at t = 0, r = 3

So, 4π(3)³ = c

C = 36π

And, at t=3, r=6

So, $\frac{4}{3}\pi {\left(6\right)}^3=3k+36\pi \left(c=36\pi \right)$

$\begin{array}{l}\Rightarrow 288\pi -36\pi =3k\\ \Rightarrow k=\frac{252\pi }{3}=84\pi \end{array}$

Hence, putting value of c and k in,

$\frac{4}{3}\pi r^3=kt+c$ , we get,

$\begin{array}{l}\frac{4}{3}\pi r^3=84\pi .t+36\pi \\ \Rightarrow r^3=\frac{3}{4\pi }\left(84\pi .t+36\pi \right)\\ \Rightarrow r^3=63t+27\\ \Rightarrow r={\left[63t+27\right]}^{\frac{1}{3}}\end{array}$

The slope of tangent is $\frac{dy}{dx}$ and slope of line joining line (-4,-3) and point say P(x,y)

$\frac{y-\left(-3\right)}{x-\left(-4\right)}=\frac{y+3}{x+4}$

So, $\frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)$

$\Rightarrow \frac{dy}{y+3}=\frac{2}{x+4}dx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{y+3}={\displaystyle \int \frac{2}{x+4}dx}}\\ \Rightarrow log\left|y+3\right|=2log\left|x+4\right|+log\left|c\right|\\ \Rightarrow log\left|y+3\right|=log{\left(x+4\right)}^2+log\left|c\right|\\ \Rightarrow log\left|y+3\right|=log\left|c{\left(x+4\right)}^2\right|\\ \Rightarrow y+3=c_1{\left(x+4\right)}^2,where,c_1=\pm c\end{array}$

Since, the curve passes through (-2,1) we get,

$\begin{array}{l}y=1,at,x=-2\\ \Rightarrow 1+3=c{\left(-2+4\right)}^2\\ \Rightarrow 4=c*4\\ \Rightarrow c=1\end{array}$

$\therefore$ The equation of the curve is $y+3={\left(x+4\right)}^2$

The slope of the tangent to then curve is $\frac{dy}{dx}$

$\begin{array}{l}\frac{dy}{dx}.y=x\\ \Rightarrow y.dy=xdx\end{array}$

So,

Integrating both sides,

$\begin{array}{l}{\displaystyle \int y.dy={\displaystyle \int xdx}}\\ \Rightarrow {\displaystyle \int \frac{y^2}{2}=\frac{x^2}{2}+c}\\ \Rightarrow y^2=x^2+A,Where,A=2c\end{array}$

As the curve passes through (0, -2) we have,

$\begin{array}{l}{\left(-2\right)}^2=0^2+A\\ \Rightarrow A=4\end{array}$

$\therefore$ The equation of the curve is

$y^2=x^2+4$

The Given D.E is

$\begin{array}{l}xy\frac{dy}{dx}=\left(x+2\right)\left(y+2\right)\\ \Rightarrow y\frac{dy}{y+2}=\frac{\left(x+2\right)}{2}dx\\ \Rightarrow \frac{y+2-2}{y+2}dy=\left(\frac{x}{x}+\frac{2}{x}\right)dx\\ \Rightarrow \left(1-\frac{2}{y+2}\right)dy=\left(1+\frac{2}{x}\right)dydx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \left(1-\frac{2}{y+2}\right)dy={\displaystyle \int \left(1+\frac{2}{x}\right)dydx}}\\ \Rightarrow y-2log\left|y+2\right|=x+2log\left|x\right|+c\\ \Rightarrow y-log{\left(y+2\right)}^2=x+log{x}^2+c\\ \Rightarrow y-x=log{\left(y+2\right)}^2+log{x}^2+c\\ \Rightarrow y-x=log\left[{\left(y+2\right)}^2.x^2\right]+c\end{array}$

A the curve passes through (-1,1) then $y=-2,at,x=1$

So, $-1-1=log{\left(-1+2\right)}^2.{\left(1\right)}^2+c$

$\begin{array}{l}\Rightarrow -2=log1+c\\ \Rightarrow c=-2\end{array}$

$\therefore$ The required equation of curve is,

$y-x=log\left[{\left(y+2\right)}^2{x}^2\right]-2$

The given D.E. is $y^1=e^{x}sinx$

$dy=e^{x}sinxdx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int e^{x}sinxdx}}\\ \Rightarrow y=I+c\end{array}$

Where, $I={\displaystyle \int e^{x}sinxdx}$

$\begin{array}{l}=sinx{\displaystyle \int e^{x}dx-{\displaystyle \int \frac{d}{dx}sinx{\displaystyle \int e^{x}dx.dx}}}\\ =sinx.e^x-{\displaystyle \int cosx{e}^{x}dx}\\ =sinx{e}^x-\left\{cosx{\displaystyle \int e^{x}dx}-{\displaystyle \int \frac{d}{dx}\left(cosx\right).{\displaystyle \int I={\displaystyle \int e^{x}x}dx}}\right\}\\ =sinx.e^x-\left\{cosx{e}^x+{\displaystyle \int sinx{e}^{x}dx}\right\}\\ =sinx.e^x-cosx{e}^x-I\\ \Rightarrow I+I=e^x\left(sinx-cosx\right)\\ \Rightarrow I=\frac{e^x}{2}\left(sinx-cosx\right)+c\end{array}$

Hence, $y=\frac{e^x}{2}\left(sinx-cosx\right)+c$

When the curve passed point (0,0),

$\begin{array}{l}y=0,at,x=0\\ \Rightarrow 0=\frac{e^x}{2}\left(sin0-cos0\right)+c\\ \Rightarrow \frac{e^0}{2}\left(0-1\right)=c\\ \Rightarrow c=\frac{1}{2}\end{array}$

$\therefore$ The required equation of the curve is $y=\frac{e^x}{2}\left(sinx-cosx\right)+\frac{1}{2}$

$\begin{array}{l}\Rightarrow 2y=e^x\left(sinx-cosx\right)+1\\ \Rightarrow 2y-1=e^x\left(sinx-cosx\right)\end{array}$