Class 12th

The given D.E is

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

$=\frac{dx}{dy}-x=y$ Which is of form $=\frac{dx}{dy}+Px=Q$

So,  $P=-1\&Q=y$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int -1dy}}=e^y$

Thus the general solution is of the form,  $x*\left(I.F\right)={\displaystyle \int Q}\left(I.F\right)dy+c$

The given D.E is

$=\frac{dy}{dx}+y\frac{\left(1+xcotx\right)}{x}=1$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=\frac{\left(1+xcotx\right)}{x}\&Q=1$

${\displaystyle \int Pdx={\displaystyle \int \left(\frac{1}{x}+\frac{xcotx}{x}\right)}}dx=log\left|x\right|+log\left|sinx\right|=log\left|xsinx\right|$

$\therefore I.F=e^{{\displaystyle \int Pdx}}-e^{{\displaystyle \int log\left|xsinx\right|}}xsinx$

Thus the solution is of the form.

$\begin{array}{l}y*xsinx={\displaystyle \int 1.xsinxdx+c}\\ =x{\displaystyle \int sinx-{\displaystyle \int \frac{d}{dx}{\displaystyle \int sinxdxdx+c}}}\\ =-2cosx+{\displaystyle \int cosxdx+c}\\ =y*xsinx=-xcosx+sinx+c\\ =y=\frac{-xcosx}{sinx}+\frac{sinx}{xsinx}+\frac{c}{xsinx}\\ =y=-cotx+\frac{1}{x}+\frac{c}{xsinx}\end{array}$

11. In semiconductors the gap between conduction band and valence band is small and hence some of the electrons from the valence band can easily jump to the conduction band and shows some conductivity but with rise in the temperature more of the electrons gets jump to the conduction band and thus, their electrical conductivity increases with rise in the temperature.

The given D.E. is

$\begin{array}{l}{\left(1+x\right)}^{2}dy+2xydx=cotxdx\\ ={\left(1+x\right)}^2\frac{dy}{dx}+2xy=cotx\end{array}$

$=\frac{dy}{dx}+\frac{2x}{1+x^2}*y=\frac{cotx}{1+x^2}$ Which is of form $\frac{dy}{dx}+Py=Q$

So , $P=\frac{2x}{1+x^2}\&Q=\frac{cotx}{1+x^2}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2x}{1+x^2}dx}}=e^{log\left|1+x^2\right|}=1+x^2$

Thus the solution is of the form,

$\begin{array}{l}\Rightarrow y\left(1+x^2\right)={\displaystyle \int \frac{cotx}{1+x^2}*}\left(1+x^2\right)dx+c\\ \Rightarrow y\left(1+x^2\right)={\displaystyle \int cotxdx+c}\end{array}$

$=log\left|sinx\right|+c$

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

139. Kindly go through the solution

The given D.E. is

$xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$

$=\frac{dy}{dx}+\frac{y}{xlogx}=\frac{2}{x^2}$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=\frac{1}{xlogx}\&Q=\frac{2}{x^2}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{x}logxdx}}=e^{{\displaystyle \int \frac{\frac{1}{x}}{logx}}dx}=e^{log\left|logx\right|}=logx$

Thus, the general solution is of the form

$\begin{array}{l}y*logx={\displaystyle \int \frac{2}{x^2}}*logxdx+c\\ y.logx=2\left[logx{\displaystyle \int \frac{1}{x^2}dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int \frac{1}{x^2}dx.dx}}}\right]+c\\ =2\left[logx*\left(\frac{x^{-1}}{-1}\right)-{\displaystyle \int \frac{1}{x}\left(\frac{x^{-1}}{-1}\right)dx}\right]+c\\ =2\left[-\frac{logx}{x}+{\displaystyle \int x^{-2}dx}\right]+c\\ =2\left[-\frac{logx}{x}-\left(\frac{x^{-1}}{-1}\right)\right]+c\end{array}$

$=ylogx=\frac{-2}{x}\left[logx+1+c\right]$

10. The ZnO crystal becomes yellow oh heating because of the metal excess defect  which is caused due to the presence of extra cations at the interstitial sites and on heating this white crystal it loses oxygen and turns yellow. The reaction involved is given as-

ZnO  Zn²⁺ + $\frac{1}{2}$ O₂ + 2^e-

Here the excess of Zn²⁺ ions move to the interstitial sites and electrons to neighbouring interstitial sites.

The given D.E. is

$x\frac{dy}{dx}+2y=x^{2}logx$

$\Rightarrow \frac{dy}{dx}+\frac{2}{x}.y=xlogx$ which is of form $\frac{dy}{dx}+Py=Q$

So, $P=\frac{2}{x}\&Q=xlogx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2}{x}dx}}=e^{2logx}=e^{log{x}^2}=x^2$

Thus, the general solution is of the form.

$y*x^2={\displaystyle \int xlogx.x^{2}dx+c}$

$={\displaystyle \int logx}{x}^{3}dx+c$

$\begin{array}{l}=logx{\displaystyle \int x^{3}dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int x^{3}dxdx+c}}}\\ =\frac{logx.x^4}{4}-{\displaystyle \int \frac{1}{4}*\frac{x^4}{4}dx+c}\end{array}$

$\begin{array}{l}=y{x}^2=\frac{x^4}{4}logx-\frac{x^4}{16}+c\\ =y=\frac{x^2}{4}logx-\frac{x^2}{16}+\frac{c}{x^2}\end{array}$

137. Yes, Let us take $f(x)=|x-1|+|x-2|.$

So, x = 1, x= 2 divides the real line into three disjoint intervals $(-\infty ,1],[1,2]$ and $[2,\infty ).$

For $x\in (-\infty ,1].$

$f(x)=-(x-1)+[-(x-2)]=-x+1-x+2=3-2x.$

For $x\in [1,2].$

$f(x)=(x-1)-(x-2)=1.$

For $x\in [2,\infty )$

$f(x)=x-1+x-2=2x-3.$

Hence, these polynomial fun are all continous and desirable. for all real values of x or, except x = 1 and x = 2.

ie, $\forall x\in R-\{1,2\}.$

For differentiavity at x = 1,

LHD = $=\underset{x\to 1^{-}}{lim}\frac{f(x)-f(1)}{x-1}=\underset{x\to 1^{-}}{lim}\frac{3-2x-1}{x-1}=\underset{x\to 1^{-}}{lim}\frac{2-2x}{x-1}.$

$=\underset{x\to 1^{-}}{lim}\frac{-2(x-1)}{x-1}$

$=\underset{x\to 1^{-}}{lim}-2$

= -2

RHD = $=\underset{x\to 1^{+}}{lim}\frac{f(x)-f(1)}{x-1}=\underset{x\to 1^{+}}{lim}\frac{1-1}{x-1}=\underset{x\to 1^{+}}{lim}\frac{0}{x-1}=0.$

as L.HD ≠ R.HD

f is not differentiable at x =1.

For continuity at x = 1.

L.HL= $=\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}=1.$

RHL = $\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}1=1$ \ LHL = RHS

f is continuous at x = 1

For continuity & differentiability at x = 2

$=\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}1=1.$

$=\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}(2x-3)=4-3=1.$

? LHL = RHL

f is continuous at x = 2

$\begin{array}{l}=\underset{x\to 2^{-}}{lim}\frac{f(x)-f(2)}{x-2}=\underset{x\to 2^{-}}{lim}\frac{1-1}{x-2}=\underset{x\to 2^{-}}{lim}=\frac{0}{x-2}\\ \end{array}$

$=\underset{x\to 2^{+}}{lim}\frac{f(x)-f(2)}{x-2}=\underset{x\to 2^{+}}{lim}\frac{2x-3-1}{x-2}$

$=\underset{x\to 2^{+}}{lim}\frac{2(x-2)}{x-2}$

$=\underset{x\to 2^{+}}{lim}2$

= 2

? LHD ≠ RHD

f is not differentiable at x = 2.

The given D.E.is

$\begin{array}{l}co{s}^{2}x\frac{dy}{dx}+y=tanx\\ =\frac{dy}{dx}+\frac{1}{co{s}^{2}x}y=\frac{tanx}{co{s}^{2}x}\end{array}$

$=\frac{dy}{dx}+se{c}^{2}xy=se{c}^{2}xtanx$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=se{c}^{2}x\&Q=se{c}^{2}xtanx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int se{c}^{2}dx}}=e^{tanx}$

Thus, the general solution is of the form.

$y.e^{tanx}={\displaystyle \int se{c}^{2}xtanx.}{e}^{tanx}dx+c$

Let, $tanx=t=se{c}^{2}xdx=dt$

$\begin{array}{l}=y{e}^t={\displaystyle \int t.}{e}^{t}dt+c\\ =t{\displaystyle \int e^{t}dt-{\displaystyle \int \frac{d}{dt}t{\displaystyle \int e^{t}dt.dt+c}}}\\ =t{e}^t-{\displaystyle \int e^{t}dt+c}\\ =t{e}^t-e^t+c\\ =e^t\left(t-1\right)+c\end{array}$

$\begin{array}{l}\Rightarrow y{e}^{tanx}=e^{tanx}\left(tanx-1\right)+c\\ y=\left(tanx-1\right)+c{e}^{-tanx}\end{array}$