Class 12th

13. In ccp lattice the number atoms per unit cell = 4 

The number of tetrahedral voids is given as = 2 (n) = 2 x 4 = 8

Only one-third of tetrahedral voids are occupied by metal M so, the ratio of atoms of element M to that of element N = 1/3 (8) : (4) = 2 : 3 

or M : N = 2 : 3 

Hence, the formula of the compound is M₂N₃.

The given D.E. is

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

$=\frac{dx}{dy}+\frac{1}{y}.x=y$ Which is of form.

$\frac{dx}{dy}+Px=Q$

So, $P=\frac{1}{y}\&Q=y$

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

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

$\begin{array}{l}x.y={\displaystyle \int y.ydy+c}\\ =xy={\displaystyle \int y^{2}dy+c}\\ =xy=\frac{y^3}{3}+c\\ =x=\frac{y^2}{3}+\frac{c}{y}\end{array}$

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.