Class 12th

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

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

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

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

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

2. A ccp structure unit cell is divided into 8 small cubes. Each small cube has atoms at alternate corners as shown in the figure. In all, each small cube has 4 atoms. When these are joined to one another, they make a regular tetrahedron. Thus, there is one tetrahedral void in each small cube and 8 tetrahedral voids in total. Each of the eight small cubes has one void in one unit cell of ccp structure. We know that ccp structure has four atoms per unit cell. Thus, the number of tetrahedral voids is twice the number of atoms.

1. When any atom is surrounded by six atoms it creates an octahedral void. In fcc, body centre is surrounded by six atoms present at the face centre. And hence one octahedral void is present at the body centre of each unit cell.

No. of octahedral void at the centre of 12 edge = 12 * $\frac{1}{4}$  = 3

No. of octahedral void at body centre = 1 

Total no. of octahedral void in ccp = 3+1+4

Location of octahedral voids per unit cell of ccp or fcc lattice (a) at body centre of the cube and (b) at the centre of each edge (only one such void is shown)

133.  Given, $cosy=xcos(a+y).$

$x=\frac{cosy}{cos(a+y)}$

Differentiating w r t 'y' we get,

$\frac{dx}{dy}=\frac{d}{dy}\left(\frac{cosy}{cos(a+y)}\right).$

$=\frac{cos(a+y)\frac{d}{dy}cosy-cosy\frac{d}{dy}cos(a+y)}{co{s}^2(a+y).}$

$=\frac{cos(a+y)(-siny)-cosy(-sin(a+y))}{co{s}^2(a+y).}$

$=\frac{-cos(a+y)siny+sin(a+y)cosy}{co{s}^2(a+y)}$

$=\frac{sin(a+y)cosy-cos(a+y)siny.}{co{s}^2(a+y)}$

$\frac{dx}{dy}=\frac{sin(a+y-y)}{co{s}^2(a+y)}\left\{\begin{array}{rr}\hfill ?sin(A-B)=sinAcosB& \hfill -cosAsinB\end{array}\right\}$

So, $\frac{dy}{dx}=\frac{co{s}^2(a+y)}{sina}$