Class 12th

8. Yellow colour in NaCl is due to the defect called as metal excess defect. In this defect anionic vacancies get created due to the diffusion of Cl-ions to the surface of the crystal and there after unpaired electrons occupy anionic sites. These sites are known as F-centres. The electrons at F-centres then absorb energy from the visible region and undergo excitation which makes the crystal appear yellow.

The given D.E. is

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

So,  $P=\frac{1}{x}=\&Q=x^2$

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

Thus, the general solution is

$\begin{array}{l}y*I.F={\displaystyle \int Q*I.Fdx+c}\\ y.x={\displaystyle \int x^2.xdx+c}\\ =xy={\displaystyle \int x^{3}dx+c}\\ =xy=\frac{x^4}{4}+c\end{array}$

7. Crystals have long range ordered arrangement of  their constituent particles but usually these crystals are not perfect as during the process of crystallisation some deviations as compared to such ideal arrangement set in depending upon the rate of cooling or presence of impurities in solution  also this process occurs at such a rate that the constituent particles may not get the sufficient time to arrange themselves in a perfect order  and hence these deviations or irregularities in arrangement is being termed as defects or imperfections. Therefore, crystals are usually not perfect. 

135. Given, $f(x)=|x{|}^3=\{\begin{array}{cc}\hfill x^3\hfill & \hfill if x\ge 0\hfill \\ \hfill -x^3\hfill & \hfill if x<0\hfill \end{array}$

For $x\ge 0,f(x)=|x{|}^3=x^3$

and $\begin{array}{cc}\hfill f^{\prime }(x)=3x^2\hfill & \hfill f^{\prime \prime }(x)=6x\hfill \end{array}$

For $x<0,f(x)=|x{|}^3={(-x)}^3=-x^3.$

so, $\begin{array}{cc}\hfill f^{\prime }(x)=-3x^2\hfill & \hfill f^{\prime \prime }(x)=-6x\hfill \end{array}$

Hence, $f^{\prime \prime }(x)=\{\begin{array}{cc}\hfill 6x,\hfill & \hfill if x\ge 0\hfill \\ \hfill -6x,\hfill & \hfill if x<0\hfill \end{array}$

Given, D.E. is

$\frac{dy}{dx}+3y=e^{-2x}$ which is of the form

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

Where $\begin{array}{l}P=3\\ \&Q=e^{-2x}\end{array}$

So, I.F $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 3dx}}=e^{3x}$

So, the solution is $=y*I.F={\displaystyle \int e^{-2x}\left(I.F\right).dx+c}$

$\begin{array}{l}=y*e^{3x}={\displaystyle \int e^{-2x}.}{e}^{3x}dx+c\\ =e^{3x}y={\displaystyle \int e^{x}dx+c}\\ =e^{3x}y=e^x+c\\ =y=\frac{e^x}{e^{3x}}+\frac{c}{e^{3x}}\\ =y=e^{-2x}+c{e}^{-3x}\end{array}$

Is the required general solution.

6. In solids the constituent particles are found to possess fixed positions and can only oscillate about their mean positions and hence they are said to be incompressible and rigid. 

The given D.E. is

$\frac{dy}{dx}+2y=2sinx$ which is of form $\frac{dy}{dx}+Px=Q$

We have, P = 2

$Q=sinx$

So, I.F. $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 2dx}}=e^{2x}$

The solution is $y*I.F={\displaystyle \int Q*\left(I.F\right)dx+c}$

$\begin{array}{l}y{e}^{2x}={\displaystyle \int sinx{e}^{2x}dx+c}\\ =y{e}^{2x}=I+c---------(1)\\ Where,I={\displaystyle \int sinx{e}^{2x}}\\ =sinx{\displaystyle \int e^{2x}-{\displaystyle \int \left(\frac{d}{dx}sinx{\displaystyle \int e^{2x}dx}\right)dx}}\\ =sinx\frac{e^{2x}}{2}-\frac{1}{2}\left[{\displaystyle \int cosx{e}^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cos{\displaystyle \int e^{2x}dx-{\displaystyle \int \left(\frac{d}{dx}cosx{\displaystyle \int e^{2x}dx}\right)dx}}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cosx\frac{e^{2x}}{2}\frac{1}{2}{\displaystyle \int \left(-cosx\right)e^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}{\displaystyle \int sinx{e}^{2x}dx}\end{array}$

$\begin{array}{l}=I=e^{2x}\frac{sinx}{2}-e^{2x}\frac{cosx}{4}-\frac{1}{4}1\\ =I+\frac{1}{4}I=2e^{2x}\frac{sinx-e^{2x}cosx}{4}\\ =\frac{5}{4}I=e^{2x}\frac{\left(2sinx-cosx\right)}{4}\\ =I=e^{2x}\frac{\left(2sinx-cosx\right)}{5}\end{array}$

Hence, equation, (1) becomes,

$\begin{array}{l}y{e}^{2x}=\frac{e^{2x}}{5}\left(2sinx-cosx\right)+c\\ =y=\frac{\left(2sinx-cosx\right)}{5}+c{e}^{2x}\end{array}$

Is the required solution.

5. The liquids and gases both are categorised as fluids as they show the ability to flow and also their molecules can move past one another freely. 

134. Given, $x=a(cost+tsint)$ and $y=a(sint-tcost).$

Differentiating w r t. 't' we get,

$\frac{dx}{dt}=a\frac{d}{dt}(cost+tsint).$

$=a\left(-sint+t\frac{d}{dt}sint+sint\frac{dt}{dt}\right).$

$=a(-sint+tcost+sint)=atcost$

$\frac{dy}{dt}=\frac{ad}{dt}(csint-tcost)$

$=a\left(cost-t\frac{d}{dt}cost-cot\frac{dt}{dt}\right)$

$=a(cost+tsint-cost)=atsint$

$\frac{dy}{dx}=\frac{dy/dt}{dx/at}=\frac{atsint}{atcost}=tant$

So, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(tant)=\frac{d}{dt}(tant)\cdot \frac{dt}{dx}.$

$=se{c}^{2}t*\frac{dt}{dx}.$

$=se{c}^{2}t*\frac{1}{(dx/dt)}$

$=se{c}^{2}t*\frac{1}{ at cost}$

$=\frac{se{c}^{3}t}{at}$