Class 12th

The integrating factor of the given differential equation  $\frac{dx}{dy}+P_{1}x=Q_1$

The general solution of the differential equation is given by,

$\begin{array}{l}x(I.F.)={\displaystyle \int \left(Q*I.F.\right)dy+C}\\ \Rightarrow x.e^{{\displaystyle \int P_{1}dy}}={\displaystyle \int \left(Q_1{e}^{{\displaystyle \int P_{1}dy}}\right)}dy+C\end{array}$

Hence, the correct answer is C.

The given differential equation is:

$\begin{array}{l}\frac{ydx-xdy}{y}=0\\ \Rightarrow \frac{ydx-xdy}{xy}=0\\ \Rightarrow \frac{1}{x}dx-\frac{1}{y}dy=0\end{array}$

Integration both sides, we get:

$\begin{array}{l}log\left|x\right|-log\left|y\right|=logk\\ \Rightarrow log\left|\frac{x}{y}\right|=logk\\ \Rightarrow \frac{x}{y}=k\\ \Rightarrow y=\frac{1}{k}x\\ \Rightarrow y=Cx,where,C=\frac{1}{k}\end{array}$

Therefore, option (C) is correct.

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$\therefore \frac{dy}{dt}\alpha y$

$\Rightarrow \frac{dy}{dt}=ky$ (k is constant)

$\Rightarrow \frac{dy}{y}=kdt$

Integration both sides, we get:

$logy=kt+C..........(1)$

In the year $1999,t=0\&y=20000.$

Therefore, we get:

$log20000=C..........(2)$

In the year $2004,t=5\&y=25000.$

Therefore, we get:

$\begin{array}{l}\Rightarrow log25000=5k+log20000\\ \Rightarrow 5k=log\left(\frac{25000}{20000}\right)=log\left(\frac{5}{4}\right)\\ \Rightarrow k=\frac{1}{5}log\left(\frac{5}{4}\right)..........(3)\end{array}$

In the year $2009,t=10years$

Now, on substituting the values of t, k, and C in equation (1), we get:

$\begin{array}{l}logy=10*\frac{1}{5}log\left(\frac{5}{4}\right)+log(20000)\\ \Rightarrow logy=log\left[20000*{\left(\frac{5}{4}\right)}^2\right]\\ \Rightarrow y=20000*\frac{5}{4}*\frac{5}{4}\\ \Rightarrow y=31250\end{array}$

Hence, the population of the village in 2009 will be 31250.

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

Integrating both sides, we get:

$\begin{array}{l}{\displaystyle \int \frac{e^{y}dy}{2-e^y}=log\left|x+1\right|+logC..........(1)}\\ Let2-e^y=t\\ \therefore \frac{d}{dy}(2-e^y)=\frac{dt}{dy}\\ \Rightarrow -e^y=\frac{dt}{dy}\\ \Rightarrow e^{y}dy=-dt\end{array}$

Substituting this value in equation (1), we get:

$\begin{array}{l}{\displaystyle \int \frac{-dt}{t}=log}og\left|x+1\right|+logC\\ \Rightarrow -log\left|t\right|=log\left|C(x+1)\right|\\ \Rightarrow -log\left|2-e^y\right|=log\left|C(x+1)\right|\\ \Rightarrow \frac{1}{2-e^y}=C(x+1)\\ \Rightarrow 2-e^y=\frac{1}{C(x+1)}..........(2)\end{array}$

Now, at x=0& y=0, equation (2) becomes:

$\begin{array}{l}\Rightarrow 2-1=\frac{1}{C}\\ \Rightarrow C=1\end{array}$

Substituting $C=1$ in equation (2), we get:

$\begin{array}{l}2-e^y=\frac{1}{x+1}\\ \Rightarrow e^y=2-\frac{1}{x+1}\\ \Rightarrow e^y=\frac{2x+2-1}{x+1}\\ \Rightarrow e^y=\frac{2x+1}{x+1}\\ \Rightarrow y=log\left|\frac{2x+1}{x+1}\right|,(x\ne -1)\end{array}$

This is the required particular solution of the given differential equation.

The given differential equation is:

$\frac{dy}{dx}+ycotx=4xcosecx$

This equation is a linear equation of the form

$\begin{array}{l}\frac{dy}{dx}+Py=Q,where,p=cotx\&Q=4xcosecx\\ Now,I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int cotxdx}}=e^{log\left|sinx\right|}=sinx\end{array}$

The general solution of the given differential equation is given by,

$y(I.F)={\displaystyle \int (Q*I.F.)dx+C}$

$\begin{array}{l}\Rightarrow ysinx={\displaystyle \int (4xcosecx.sinx)dx+C}\\ \Rightarrow ysinx=4{\displaystyle \int xdx+C}\\ \Rightarrow ysinx=4.\frac{x^2}{2}+C\\ \Rightarrow ysinx=2x^2+C..........(1)\\ Now,y=0at,x=\frac{\pi }{2}\end{array}$

Therefore, equation (1) becomes:

$\begin{array}{l}0=2*\frac{\pi }{2}+C\\ \Rightarrow C=-\frac{{\pi }^2}{2}\end{array}$

Substituting $C=-\frac{{\pi }^2}{2}$ in equation (1), we get:

$ysinx=2x^2-\frac{{\pi }^2}{2}$

This is the required particular solution of the given differential equation.

Kindly go through the solution

$\begin{array}{l}(x-y)(dx+dy)=dx-dy\\ \Rightarrow (x-y+1)dy=(1-x+y)dx\\ \Rightarrow \frac{dy}{dx}=\frac{1-x+y}{x-y+1}\\ \Rightarrow \frac{dy}{dx}=\frac{1-(x-y)}{1+(x-y)}..........(1)\\ Let,x-y=t\\ \Rightarrow \frac{d}{dx}(x-y)=\frac{dt}{dx}\\ \Rightarrow 1-\frac{dy}{dx}=\frac{dt}{dx}\\ \Rightarrow 1-\frac{dt}{dx}=\frac{dy}{dx}\end{array}$

Substituting the values of $x-y$ and $\frac{dy}{dx}$ in equation (1), we get:

$\begin{array}{l}1-\frac{dt}{dx}=\frac{1-t}{1+t}\\ \Rightarrow \frac{dt}{dx}=1-\left(\frac{1-t}{1+t}\right)\\ \Rightarrow \frac{dt}{dx}=\frac{(1+t)-(1-t)}{1+t}\\ \Rightarrow \frac{dt}{dx}=\frac{2t}{1+t}\end{array}$

$\begin{array}{l}\Rightarrow \left(\frac{1+t}{t}dt\right)=2dx\\ \Rightarrow \left(1+\frac{1}{t}\right)dt=2dx..........(2)\end{array}$

Integrating both sides, we get:

$\begin{array}{l}t+log\left|t\right|=2x+C\\ \Rightarrow (x-y)+log\left|x-y\right|=2x+C\\ \Rightarrow log\left|x-y\right|=x+y+C..........(3)\end{array}$

$Now,y=-1,at,x=0$

Therefore, equation (3) becomes:

$log1=0-1+C$

$\Rightarrow C=1$

Substituting $C=1$ in equation (3), we get:

$og\left|x-y\right|=x+y+1$

This is the required particular solution of the given differential equation .

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

$Let,e^{\frac{x}{y}}=z$

Differentiating it with respect to y, we get:

$\begin{array}{l}\left(e^{\frac{x}{y}}\right)=\frac{dz}{dy}\\ \Rightarrow e^{\frac{x}{y}}.\frac{d}{dy}\left(\frac{x}{y}\right)=\frac{dz}{dy}\\ \Rightarrow e^{\frac{x}{y}}.\left[\frac{y.\frac{dx}{dy}-x}{y^2}\right]=\frac{dz}{dy}..........(2)\end{array}$

From equation (1) and equation (2), we get:

$\begin{array}{l}\frac{dz}{dy}=1\\ \Rightarrow dz=dy\end{array}$

Integration both sides, we get:

$\begin{array}{l}z=y+C\\ \Rightarrow e^{\frac{x}{y}}y+C\end{array}$