Maths

20. LetP(n): $10^{2n-1}+$ 1 is divisible by 11.

Putting n = 1

$P(1)=10^{\prime }+1=11$ is divisible by 11.

Which is true. Thus, P(1) is true.

Let us assume that P(k) is true for some natural no. k.

P(k)= $10^{2k-1}+$

$\mathrm{}\mathrm{1\; is\; divisible\; by\; 11.}$

$10^{2k-1}+1=11aa\in z$

$\to 10^{2k-1}=11a-1$ (1)

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

$P(k+1):10^{2(K+1)-1}+1=10^{2k+1}+1\mathrm{is\; divisible\; by\; 11.}$

$10^{2k+1}+1=10^{(2k-1)+2}+1$

$=10^{2k-1}\cdot 10^2+1$

$=100(11a-1)+1(using(1))$

=1100a $-$ 99= 11(100a $-$ 9)

11b where b= (100a $-$ 9) $\in z$

$\Rightarrow 1{0^2}^{k+1}+1$ is divisible by 11.

$\Rightarrow P(k+1)$

 is true when p(k) is true.

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

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.

97. Let $y=xcosx$

So,  $\frac{dy}{dx}=x\frac{d}{dx}cosx+\frac{dx}{dx}cosx$

$=-xsinx+cosx$

$\therefore \frac{d^{2}y}{d{x}^2}=-x\frac{d}{dx}sinx-sinx\frac{dx}{dx}+\frac{d}{dx}cosx$

$=-xcosx-sinx+\left(-sinx\right)$

$=-\left(xcosx+2sinx\right)$

Kindly check the Answer:

The equation of the family of ellipses having foci on the y-axis and the centre at origin is as follows:

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

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

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

Again, differentiating with respect to x, we get:

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

Substituting this value in equation (2), we get:

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

This is the required differential equation

The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:

$x^2 =4ay$

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

$2x=4ay\text{'}$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l}\frac{2x}{x^2}=\frac{4a{y}^{\text{'}}}{4ay}\\ \Rightarrow \frac{2}{x}=\frac{y^{\text{'}}}{y}\\ \Rightarrow x{y}^{\text{'}}=2y\\ \Rightarrow x{y}^{\text{'}}-2y=0\end{array}$

This is the required differential equation.

96. Let $y=x^{20}$

So,  $\frac{dy}{dx}=20x^{20-1}=20x^{19}$

$\therefore \frac{d^{2}y}{d{x}^2}=20*19x^{19-1}$

$=380x^{18}$