Maths

104. Let $y=sin\left(logx\right)$

so, $\frac{dy}{dx}=\frac{d}{dx}sin\left(logx\right)=cos\left(logx\right)\frac{d}{dx}logx=\frac{cos\left(logx\right)}{x}$

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

$=\frac{x\left[-sin\left(logx\right)\right]\frac{d}{dx}logx-cos\left(logx\right)}{x^2}$

$=\frac{-\left[x\cdot sin\left(logx\right)*\frac{1}{x}+cos\left(logx\right)\right]}{x^2}$

$=\frac{-\left[sin\left(logx\right)+cos\left(logx\right)\right]}{x^2}$

Given,  $\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)$

$\Rightarrow \frac{dy}{\left(1+y^2\right)}=\left(1+x^2\right)dx$

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(1+y^2\right)}dy={\displaystyle \int \left(x^2+1\right)dx}}\\ \Rightarrow ta{n}^{-1}\frac{y}{1}=\frac{x^3}{3}+x+c\end{array}$

$\Rightarrow ta{n}^{-1}y=\frac{x^3}{3}+x+c$ is the general solution.

103. Let $y=log\left(logx\right)$

So,  $\frac{dy}{dx}=\frac{1}{logx}\frac{d}{dx}logx=\frac{1}{xlogx}$

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

$=\frac{-\left[x\frac{d}{dx}logx+logx\frac{dx}{dx}\right]}{{\left[xlogx\right]}^2}$

$=\frac{-\left(x*\frac{1}{x}+logx\right)}{{\left[xlogx\right]}^2}$

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

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

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

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

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

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

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

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

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