Maths

Given,  $\left(e^x+e^{-x}\right)dy-\left(e^x-e^{-x}\right)dx=0$

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

Integrating both sides

$\Rightarrow {\displaystyle \int dy=}\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\left\{?{\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx=log\left|x\right|\right\}$

$\Rightarrow y=log\left|e^x+e^{-x}\right|+c$ is the required general solution.

Given, $se{c}^{2}xtanydx+se{c}^{2}ytanxdy=0$

Dividing throughout by ' $tanxtany$ ' we get,

$\begin{array}{l}\frac{se{c}^{2}xtany}{tanxtany}dx+\frac{se{c}^{2}ytanx}{tanxtany}dy=0\\ \Rightarrow \frac{se{c}^{2}x}{tanx}dx+\frac{se{c}^{2}y}{tany}dy=0\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{se{c}^{2}x}{tanx}dx+{\displaystyle \int \frac{se{c}^{2}y}{tany}dy=logc}}\\ \Rightarrow log\left|tanx\right|+log\left|tany\right|=logc\left\{{\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx-log\left|f\left(x\right)\right|\right\}\\ \Rightarrow log\left|\left(tanx+tany\right)\right|=logc\end{array}$

$\Rightarrow tanxtany=\pm c$ is the required general solution.

102. Let $y=ta{n}^{-1}x$

So,  $\frac{dy}{dx}=\frac{d}{dx}ta{n}^{-1}x=\frac{1}{1+x^2}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{\left(1+x^2\right)\frac{d}{dx}\left(1\right)-\left(1\right)\frac{d}{dx}\left(1+x^2\right)}{{\left(1+x^2\right)}^2}$

$=\frac{-2x}{{\left(1+x^2\right)}^2}$

Given,  $\frac{dy}{dx}+y=1$

$\Rightarrow \frac{dy}{dx}=1-y=-\left(y-1\right)$

By separable of variable,

$\frac{dy}{\left(y-1\right)}=-dx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(y-1\right)}=-{\displaystyle \int dx}}\\ \Rightarrow log\left|y-1\right|=-x+c\\ \Rightarrow \left|y-1\right|=e^{-x+c}\\ \Rightarrow y-1=\pm e^{-x}.e^c\end{array}$

$\Rightarrow y=1+Ac$ where $A=\pm e^c$

Is the general solution.

Kindly go through the solution

24. Let P (n) be the statement “ 2n+7< (n+3)2

ofn=1

P (1): 2 $*1+7<{(1+3)}^2$

$9<4^2$

9<16 which is true. This P (1) is true.

Suppose P (k) is true.

P (k)= 2k+7< (k+3)^{2   . (1)}              

Lets prove that P (k +1) is also true.

“ 2 (k + 1) + 7 < (k + 4)²=k²+ 8k + 16”

P (k +1) = 2 (k +1) +7 = (2k +7) +2

  < (k +3)²+ 2  (Using 1)

= k²+ 9 + 6k +2 = k²+6k +11

Adding and subtracting (2k + k) in the R. H. S.

$=k^2+6k+11+2k+5-(2k-5)$

$=\left(k^2-8k+16\right)-(2k-5)$

$={(k+4)}^2-(2k-5)$

$<{(k+4)}^2\mathrm{,\; since\; 2}k+5>0$ for all$k\in N$

$\Rightarrow P(k+1)$ is true.

$\therefore$ By the principle of mathematical induction, P (n)is true for all n $\in$ N.

101. Let $y=e^{6x}cos3x$

So, $\frac{dy}{dx}=e^{6x}\frac{d}{dx}cos3x+cos3x\frac{d}{dx}{e}^{6x}$

$=e^{6x}\left(-sin3x\right)\frac{d}{dx}\left(3x\right)+cos3x\cdot e^{6x}\frac{d}{dx}\left(6x\right)$

$=e^{6x}\left[-3sin3x+6cos3x\right]$

$\therefore \frac{d^{2}y}{d{x}^2}=e^{6x}\frac{d}{dx}\left[-3sin3x+6cos3x\right]+\left[-3sin3x+6cos3x\right]\frac{d}{dx}{e}^{6x}$

$=e^{6x}\left[-3cos3x\frac{d}{dx}\left(3x\right)+6\left(-sin3x\right)\frac{d}{dx}\left(3x\right)\right]+\left[-3sin3x+6cos3x\right]e^{6x}\frac{d}{dx}\left(6x\right)$

$=e^{6x}\left\{-9cos3x-18sin3x-18sin3x+36cos3x\right\}$

$=e^{6x}\left(27cos3x-36sin3x\right)$

$=9e^{6x}\left(3cos3x-4sin3x\right)$

The given D.E is

$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

By separable of variable,

$\begin{array}{l}\Rightarrow dy=\frac{1-cosx}{1+cosx}dx\left\{\begin{array}{l}cos2x=1-2si{n}^{2}x\\ =2si{n}^{2}x=1-cos2x\\ =2si{n}^2\frac{x}{2}=1-cosx\\ cos2x=2co{s}^{2}x-1\end{array}\right\}\\ \Rightarrow dy=\frac{2si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}dx\\ \Rightarrow dy=ta{n}^2\frac{x}{2}dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int ta{n}^2\frac{x}{2}dx\left\{se{c}^{2}x=1+ta{n}^2\right\}}}\\ \Rightarrow y={\displaystyle \int \left(se{c}^2\frac{x}{2}-1\right)}dx\end{array}$

$\Rightarrow y=\frac{tan\frac{x}{2}}{\frac{1}{2}}-x+c$ c = constant

$\Rightarrow y=2tan\frac{x}{2}-x+c$ is the general solution.

23. Let $P(n):41^n-14^n\mathrm{is\; a\; multiple\; of\; 27.}$

Put n= 1,

$P(1)=41-14=27$ is a multiple of 27.

Which is true.

Assume that P(k) is true for some natural no. k.

P(k)= $41^k-14^k$ be a multiple of 27

i.e, $41^k-14^k=27a,a\in z$

$\Rightarrow 41^k=27a+14^k$ (1)

We want to prove that P(k+1) is also true.

Now,

$P(k+1)=41^{k+1}-14^{k+1}$

$=41^k\cdot 41-14\cdot 14^k$

$=41\left(27a+14^k\right)-14\cdot 14^k$ (Using 1)

$=41*27a+41\cdot 14^k-14\cdot 14^k$

$=-27\left(24+a+14^k\right)=41*27a+14^k(41-14)$

$=41*27a+27\cdot 14^k$

$=27\left(41a+14^k\right)$

$=27b,whereb=\left(41a+14^k\right)\in z$

$\Rightarrow 41^{k+1}-14^{k+1}\mathrm{is\; multiple\; of\; 2}7$

$\Rightarrow P(k+1)$ is true when P(k) is true.

Hence, by P.M.I. P(n) is true for every positive integer n.

100. Let $y=e^{x}sin5x$

so, $\frac{dy}{dx}=e^x\frac{d}{dx}sin5x+sin5x\frac{d}{dx}{e}^x$

$=e^x\cdot cos5x\frac{d}{dx}\left(5x\right)+e^{x}sin5x$

$=5e^{x}cos5x+e^{x}sin5x.$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(5e^{x}cos5x+e^{x}sin5x\right)$

$=5e^x\frac{d}{dx}cos5x+5cos5x\frac{d}{dx}{e}^x+e^x\frac{d}{dx}sin5x+sin5x\frac{d}{dx}{e}^x$

$=-5e^{x}sin5x\frac{d}{dx}\left(5x\right)+5e^{x}cos5x+e^{x}cos5x\frac{d}{dx}\left(5x\right)+e^{x}sin5x$

$=-25e^{x}sin5x+5e^{x}cos5x+5e^{x}cos5x+e^{x}sin5x$

$=e^x\left(10\cdot cos5x-24sin5x\right)$

$=2e^x\left(5cos5x-12sin5x\right)$