Ncert Solutions Maths class 12th

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

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:

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

$\begin{array}{l}\left(1+e^{2x}\right)dy+\left(1+y^2\right)e^{x}dx=0\\ \Rightarrow \frac{dy}{1+y^2}+\frac{e^{x}dx}{1+e^{2x}}=0\end{array}$

Integrating both sides, we get:

$\begin{array}{l}ta{n}^{-1}y+{\displaystyle \int \frac{e^{x}dx}{1+e^{2x}}=C..........(1)}\\ Let,e^x=t\Rightarrow e^{2x}=t^2\\ \Rightarrow \frac{d}{dx}\left(e^x\right)=\frac{dt}{dx}\\ \Rightarrow e^x=\frac{dt}{dx}\\ \Rightarrow e^{x}dx=dt\end{array}$

Substituting these values in equation (1), we get:

$\begin{array}{l}ta{n}^{-1}y+{\displaystyle \int \frac{dt}{1+t^2}}=C\\ \Rightarrow ta{n}^{-1}y+ta{n}^{-1}t=C\\ \Rightarrow ta{n}^{-1}y+ta{n}^{-1}(e^x)=C..........(2)\\ Now,y=1,at,x=0\end{array}$

Therefore, equation (2) becomes:

$\begin{array}{l}ta{n}^{-1}1+ta{n}^{-1}1=C\\ \Rightarrow \frac{\pi }{4}+\frac{\pi }{4}=C\\ \Rightarrow C=\frac{\pi }{2}\end{array}$

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

$ta{n}^{-1}y+ta{n}^{-1}(e^x)=\frac{\pi }{2}$

This is the required solution of the given differential equation.

The differential equation of the given curve is:

$\begin{array}{l}sinxcosydx+cosxsinydy=0\\ \Rightarrow \frac{sinxcosydx+cosxsinydy}{cosxcosy}=0\\ \Rightarrow tanxdx+tanydy=0\end{array}$

Integrating both sides, we get:

$\begin{array}{l}log(secx)+log(secy)=logC\\ log(secx.secy)=logC\\ \Rightarrow secx.secy=C..........(1)\end{array}$

The curve passes through point $\left(0,\frac{\pi }{4}\right)$

$\begin{array}{l}\therefore 1*\surd 2=C\\ \Rightarrow C=\surd 2\end{array}$

On subtracting $C=\surd 2$ in equation (10, we get:

$\begin{array}{l}secx.secy=\surd 2\\ \Rightarrow secx.\frac{1}{cosy}=\surd 2\\ \Rightarrow cosy=\frac{sec\mathrm{x/\surd 2}}{}\end{array}$

Given: Differential equation $\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0$

$\begin{array}{l}\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0\\ \Rightarrow \frac{dy}{dx}=-\frac{(y^2+y+1)}{x^2+x+1}\\ \Rightarrow \frac{dy}{y^2+y+1}=\frac{-dx}{x^2+x+1}\\ \Rightarrow \frac{dy}{y^2+y+1}+\frac{dx}{x^2+x+1}=0\end{array}$

Integrating both sides,

Kindly go through the solution

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

${(x-a)}^2+{(y-a)}^2=a^2..........(1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2(x-a)+2(y-a)\frac{dy}{dx}=0\\ \Rightarrow (x-a)+(y-a)y^{\text{'}}=0\\ \Rightarrow x-a+y{y}^{\text{'}}-a{y}^{\text{'}}=0\\ \Rightarrow x+y{y}^{\text{'}}-a(1+y^{\text{'}})=0\\ \Rightarrow a=\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\end{array}$

Substituting the value of a in equation (1), we get:

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

Hence, the required differential equation of the family of circles is ${(x-y)}^2\left[1+(y^{\text{'}}){}^2\right]=(x+y{y}^{\text{'}}){}^2$