Differential Equations

Kindly go through the solution

Integrating both sides,

The given D.E. is

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

Hence, the D.E. is homogenous $f{x}^n$

Let, $y=vx.=\frac{y}{x}=v$ so that, $\frac{dy}{dx}=v+x\frac{dv}{dx}$ is the D.E.

Thus, $v+x\frac{dv}{dx}=1-2v^2+v$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=1-2v^2\\ \Rightarrow \frac{dv}{1-2v^2}=\frac{dx}{x}\end{array}$

Integrating both sides,

The Given D.E. is

$\begin{array}{l}\left(x^2-y^2\right)dx+2xydy=0\\ \Rightarrow 2*ydy=-\left(x^2-y^2\right)dx\\ =\frac{dy}{dx}=-\frac{\left(x^2-y^2\right)}{2xy}=\frac{\left(y^2-x^2\right)}{2xy}\\ =\frac{1}{2}\frac{\left(\frac{y^2-x^2}{x^2}\right)}{\left(\frac{xy}{x^2}\right)}\\ =\frac{1}{2}\frac{\left[{\left(\frac{y}{x}\right)}^2-1\right]}{\left(\frac{y}{x}\right)}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E. is homogenous.

Let, $y=vx\Rightarrow \frac{y}{x}=v,sothat,\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

$\begin{array}{l}\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}\left[\frac{v^2-1}{v}\right]\\ \Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{v^2-1-2v^2}{2v}=\frac{-1-v^2}{2v}\\ \Rightarrow \left(\frac{2v}{1+v^2}\right)dv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

Putting back $v=\frac{y}{x}$ we get,

$\begin{array}{l}\Rightarrow x\left[\frac{y^2}{x^2}+1\right]=c\\ \Rightarrow \frac{y^2+x^2}{x}=c\end{array}$

$\Rightarrow y^2+x^2=xc$ is the required solution.

The Given D.E. is $\left(x-y\right)dy-\left(x+y\right)dx=0$

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

Hence, the given D.E. is homogenous.

Let, $y=vx=\frac{y}{x}=v,so,\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

Then, $\begin{array}{l}v+x\frac{dv}{dx}=\frac{1+v}{1-v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-1+v^2}{1-v}=\frac{1+v^2}{1-v}\\ \Rightarrow \left[\frac{1-v}{1+v^2}\right]dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{1-v}{1+v^2}dv={\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int \frac{1}{1+v^2}dv-\frac{1}{2}{\displaystyle \int \frac{2v}{1+v^2}dv=logx+c}}\\ \Rightarrow ta{n}^{-1}v-\frac{1}{2}log\left(1+v^2\right)=logx+c\end{array}$

Putting back $v=\frac{y}{x},weget,$

$\begin{array}{l}=ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left|1+\frac{y^2}{x^2}\right|=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left|\frac{x^2+y^2}{x^2}\right|=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}\left[log\left(x^2+y^2\right)-log{x}^2\right]=logx+c\end{array}$

$\begin{array}{l}=ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+\frac{1}{2}log{x}^2=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+log{\left(x^2\right)}^{\frac{1}{2}}=logx+c\\ =ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log\left(x^2+y^2\right)+logx=log+c\\ =ta{n}^{-1}\frac{y}{x}=\frac{1}{2}log\left(x^2+y^2\right)+c\end{array}$

The given D.E. is

$\begin{array}{l}{y}^1=\frac{x+y}{x}\\ =\frac{dy}{dx}=1+\frac{y}{x}=f\left(\frac{y}{x}\right)\end{array}$

Hence, the given D.E is homogenous.

Let,  $y=vx\to \frac{y}{x}=v,sothat,\frac{dy}{dx}=v+\frac{xdv}{dx}$

So, the D.E. becomes

$\begin{array}{l}v+\frac{xdv}{dx}=1+v\\ \Rightarrow \frac{xdv}{dx}=1\\ \Rightarrow dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dv={\displaystyle \int \frac{dx}{x}}}\\ v=log\left|x\right|+c\end{array}$

Putting $v=\frac{y}{x}$ back we get,

$\frac{y}{x}=log\left|x\right|+c$

The given D.E. is

$\begin{array}{l}\frac{dy}{dx}=\frac{x^2+y^2}{x^2+x+y}=\frac{x^2\left(1+\frac{y^2}{x^2}\right)}{x^2\left(1+\frac{y}{x}\right)}=F\left(x,y\right)\\ Then,F\left(\lambda x,\lambda y\right)=\frac{{\lambda }^2{x}^2}{{\lambda }^2{x}^2}\left[\frac{1+\frac{{\lambda }^2{y}^2}{{\lambda }^2{x}^2}}{1+\frac{\lambda y}{\lambda x}}\right]={\lambda }^{0}F\left(x,y\right)\\ ={\lambda }^{2}F\left(x,y\right)\end{array}$

Hence, $F\left(x,y\right)$ is a homogenous $f{x}^n$ of degree 2.

To solve it, let

$\begin{array}{l}y=vx,sothat,\frac{dy}{dx}=v.\frac{dx}{dx}+x\frac{dv}{dx}\\ v=\frac{y}{x}\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$

The D.E. now becomes,

$\begin{array}{l}v+x\frac{dv}{dx}=\frac{1+v^2}{1+v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1+v}-v=\frac{1+v^2-v-v^2}{1+v}=\frac{1-v}{1+v}\\ \Rightarrow \left(\frac{1+v}{1-v}\right)dv=\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{1+v}{1-v}dv={\displaystyle \int \frac{dx}{x}}}\\ \Rightarrow {\displaystyle \int \frac{1+v}{1-v}dv=logx+c}\\ \Rightarrow {\displaystyle \int \frac{2-\left(1-v\right)}{1-v}dv=logx+c}\\ \Rightarrow {\displaystyle \int \frac{2}{1-v}dv-{\displaystyle \int 1dv=logx+c}}\\ \Rightarrow \frac{2log\left|1-v\right|}{-1}-v=logx+c\\ \Rightarrow log{\left(1-v\right)}^2+v=-logx-c\end{array}$

Put $v=\frac{y}{x},$

$\begin{array}{l}log{\left(1-\frac{y}{x}\right)}^2+\frac{y}{2}=-logx+c\\ \Rightarrow log{\left(\frac{x-y}{x}\right)}^2+logx=-\frac{y}{x}-c\end{array}$

$\begin{array}{l}\Rightarrow log\left[{\left(\frac{x-y}{x}\right)}^2*x\right]=-\frac{y}{x}-c\\ \Rightarrow \frac{x-y}{x}=e^{-\frac{y}{x}-c}=c_1{e}^{-\frac{y}{x}-c},where,c_{1}e\end{array}$

$\Rightarrow {\left(x-y\right)}^2=c_1.x{e}^{-\frac{y}{x}}$ is the required solution of the D.E.

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, $100=e^{\frac{r*0}{100}+c}$

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