Ncert Solutions Maths class 12th

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.

100. Let $y=e^{x}sin5x$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=2x+6xlogx+3x$

$=5x+6xlogx$

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.

Let the centre of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

$\begin{array}{l}{x}^2+{\left(y-b\right)}^2=3^2\\ \Rightarrow x^2+{\left(y-b\right)}^2=9..........(1)\end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2x+2(y-b).y^{\text{'}}=0\\ \Rightarrow (y-b).y^{\text{'}}=-x\\ \Rightarrow y-b=\frac{-x}{y^{\text{'}}}\end{array}$

Substituting the value of $(y-b)$ in equation (1), we get:

$\begin{array}{l}{x}^2+{\left(\frac{-x}{y^{\text{'}}}\right)}^2=9\\ \Rightarrow x^2\left[1+\frac{1}{{\left(y^{\text{'}}\right)}^2}\right]=9\\ \Rightarrow x^2\left({\left(y^{\text{'}}\right)}^2+1\right)=9{\left(y^{\text{'}}\right)}^2\\ \Rightarrow (x^2-9){\left(y^{\text{'}}\right)}^2+x^2=0\end{array}$

This is the required differential equation.

The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1..........(1)$

Differentiating both sides of equation (1) with respect to x, we get:

$\begin{array}{l}\frac{2x}{a^2}-\frac{2y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow \frac{x}{a^2}-\frac{y{y}^{\text{'}}}{b^2}=0..........(2)\end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l}\frac{1}{a^2}-\frac{y^{\text{'}}.y^{\text{'}}+y.y^{"}}{b^2}=0\\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)=0\end{array}$

Substituting the value of $\frac{1}{a^2}$ in equation (2), we get:

$\begin{array}{l}\frac{x}{b^2}\left({\left(y^{\text{'}}\right)}^2+y{y}^{"}\right)-\frac{y{y}^{\text{'}}}{b^2}=0\\ \Rightarrow x{\left(y^{\text{'}}\right)}^2+xy{y}^{"}-y{y}^{\text{'}}=0\\ \Rightarrow xy{y}^{"}+x{\left(y^{\text{'}}\right)}^2-y{y}^{\text{'}}=0\end{array}$

This is the required differential equation.