Maths

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

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

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

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

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

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

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

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

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

22. Let P(n): $3^{2n+2}-8n-9$ is divisible by 8

put n= 1,

P(1): $3^{2+2}-8.1-9$

3⁴ – 8 – 9 = 81– 17 = 64= is divisible by 8

Which is true.

Assume that P(k) is true for some natural numbers k.

i.e, $3^{2k+2}-8k-9$ be divisible by 8

$3^{2k+2}-8k-9=8a$ where,a $\in z$

$3^{2k+2}=8a+8k+9$ (1)

We want to prove thatP(k+ 1) is true.

$P(k+1):3^{2(k+1)+2}-8(k+1)-9$ is divisible by 8, is also true.

Now,

${3^{2k+2+}}^2-=8(k+1)-9$

= $3^{2k+4}-8k+8-9$

$3^{(2k+2)+2}-8k+8-9$

3^{(2k +2)}. 3² $-$ 8k $-$ 17

$=9(8a+8k+9)-8k-17$ (Using 1)

$=72a+72k+81-8k-17$

= 72a + 64k+ 64 = 8(9a + 8k + 8)

= 8b, where b = 9a + 8b + 8 $a\in z$

$\Rightarrow$ 3^{2k + 4}– 8(k+1) – 9 is divisible by 8.

$\therefore$ P(k+1) is true when P(k) is true. Hence, By P.M.I. P(n) is true for all positive integer n.

The highest order derivation present in the D.E. is y, so its order is 1.

As the given D.E. is a polynomial equation in its derivative its degree is 1.

99. Let $y=x^{3}logx$

So,  $\frac{dy}{dx}=x^3\frac{d}{dx}logx+log2.\frac{d{x}^3}{dx}$

$=x^3.\frac{1}{x}+logx\cdot 3x^2$

$=x^2+logx\left(3x^2\right)$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(x^2+logx\cdot 3x^2\right)$

$=2x+6xlogx+3x^2*\frac{1}{x}$

$=2x+6xlogx+3x$

$=5x+6xlogx$

21. Let 

$\mathrm{Putting\; x}=1,$

$P(1)=x^2-y^2\mathrm{is\; divisible\; by\; x}+y$ or

$\left(x+y\right)(x-y)isdivisiblebyx+y,$ which is true.

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

$P(k)=x^{2k}-y^{2k}$ is divisible by x + y

i.e. $x^{2k}-y^{2k}=a(x+y)$ where $\in z$

$\Rightarrow x^{2k}=a(x+y)+y^{2k}$ (1)

Now, let us prove P(k +1) is true.

$P(k+1):x^{2(k+1)}-y^{2(k+1)}$

$=x^{2x+2}-y^{2x+2}$

$=x^2\cdot x^{2k}-y^2\cdot y^{2k}$

$=x^2\left[a(x+y)+y^{2k}\right]-y^2\cdot y^{2k}[using(1)$
$]$

$=a{x}^2(x+y)+x^2\cdot y^{2k}-y^2\cdot y^{2k}$

$=a{x}^2(x+y)+y^{2k}\left(x^2-y^2\right)$

$=a{x}^2(x+y)+y^{2k}(x+y)(x-y)$

$=(x+y)\left[a{x}^2+y^{2k}(x-y)\right]$

$=b(x+y)whereb=\left[a{x}^2+y^{2k}(x-y)\right]\in z$

$\Rightarrow x^{2k+2}-y^{2k+2}\mathrm{is\; divisible\; by\; x}+y$

$\therefore$ P(k+ 1) is true where P (k) is true.

Hence, by P.M.I. P(n) is true for all natural number i.e.

The given equation of curve is  $y = x$ .

Differentiating with respect to x, we get:

$\frac{dy}{dx}=1 ..........(1)$

Again, differentiating with respect to x, we get:

$\frac{d^{2}y}{d{x}^2}=0..........(2)$

Now, on substituting the values of y,  $\frac{d^{2}y}{d{x}^2}$ and $\frac{dy}{dx}$   from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}-x^2\frac{dy}{dx}+xy=0-x^2.1+x.x\\ =-x^2+x^2\\ =0\end{array}$

Therefore, option (C) is correct.

98. Let $y=logx$

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

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}{x}^{-1}=-1x^{-1-1}=\frac{-1}{x^2}$

Given: $y=c_1{e}^x+c_2{e}^{-x}.........(1)$

Differentiating with respect to x, we get:

$\frac{dy}{dx}=c_1{e}^x-c_2{e}^{-x}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=c_1{e}^x+c_2{e}^{-x}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=y\\ \Rightarrow \frac{d^{2}y}{d{x}^2}-y=0\end{array}$

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.