Maths

The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

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

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

$\begin{array}{l}2x+2y{y}^{\text{'}}=2a\\ \Rightarrow x+y{y}^{\text{'}}=a\end{array}$

Now, on substituting the value of a in equation (1), we get:

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

This is the required differential equation.

Given: Equation of the family of curves $y=e^x\left(acosx+bsinx\right)..........(i)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}=e^x\left(acosx+bsinx\right)+e^x\left(-acosx+bsinx\right)\\ \Rightarrow y^{\text{'}}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]..........(2)\end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l}{y}^{"}=e^x\left[\left(a+b\right)cosx-\left(a-b\right)sinx\right]+e^x\left[-\left(a+b\right)sinx-\left(a-b\right)cosx\right]\\ y^{"}=e^x\left[2bcosx-2asinx\right]\\ y^{"}=2e^x\left(bcosx-asinx\right)\\ \Rightarrow \frac{y^{"}}{2}=e^x\left(bcosx-asinx\right)..........(3)\end{array}$

Adding equations (1) and (3), we get:

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

This is the required differential equation of the given curve.

95. Let y = x2 + 3x + 2

So,  $\frac{dy}{dx}=2x+3+0$  (differentiation w r t 'x')

$?\frac{d^{2}y}{d{x}^2}=2+0$  (Again “ “ ) = 2

17. We can write the given statement as:-

$\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+?+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}$

For n = 1,

We get $P(1)=\frac{1}{3\cdot 5}=\frac{1}{15}=\frac{1}{3(2\cdot 1+3)}=\frac{1}{3(2+3)}=\frac{1}{3*5}=\frac{1}{15}$

Which is true.

Consider P(k) be true for some positive integer k.

$P(K)=\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+\dots +\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)}$ (1)

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

Now,

P(k +1) = $1$ $\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}+\frac{1}{[2(k+1)+1][2(k+1)+3]}$

By using (1),

$\begin{array}{l}=\frac{k}{3(2k+3)}+\frac{1}{(2k+2+1)(2k+2+3)}\\ =\frac{k}{3(2k+3)}+\frac{1}{(2k+3)(2k+5)}\\ =\frac{1}{(2k+3)}\left[\frac{k}{3}+\frac{1}{(2k+5)}\right]\\ =\frac{1}{(2k+3)}\left[\frac{k(2k+5)+3}{3(2k+5)}\right]\end{array}$

$=\frac{1}{(2k+3)}\left[\frac{2k^2+5k+3}{3(2k+5)}\right]$

= $\frac{1}{2k+3}\left[\frac{2k^2+2k+3k+3}{3(2k+5)}\right]$

$=\frac{1}{2k+3}\{\frac{2k(k+1)+3(k+1)}{3(2k+1)}\}$

$\begin{array}{l}=\frac{1}{(2k+3)}\frac{(2k+3(k+1)}{\cdot (2k+1)\cdot 3}\\ =\frac{k+1}{3(2k+5)}\\ =\frac{(k+1)}{3\{2(k+1)+3\}}\end{array}$

P(k+ 1) is true wheneverP(k) is true.

Therefore, from the principle of mathematical induction, theP(n) is true for all natural number n.

Given: Equation of the family of curves  $y=e^{2x}\left(a+bx\right)..........(1)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}{y}^{\text{'}}=2e^{2x}\left(a+bx\right)+e^{2x}.b\\ \Rightarrow y^{\text{'}}=e^{2x}\left(2a+2bx+b\right)..........(2)\end{array}$

Multiplying equation (1) with (2) and then subtracting it from equation (2), we get:

$\begin{array}{l}{y}^{\text{'}}-2y=e^{2x}\left(2a+2bx+b\right)-e^{2x}\left(2a+2bx\right)\\ \Rightarrow y^{\text{'}}-2y=b{e}^{2x}..........(3)\end{array}$

Differentiating both sides with respect to x, we get:

$y^{"}-2y^{\text{'}}=2b{e}^{2x}..........(4)$

Dividing equation (4) by equation (3), we get:

$\begin{array}{l}\frac{y^{"}-2y^{\text{'}}}{y^{\text{'}}-2y}=2\\ \Rightarrow y^{"}-2y^{\text{'}}=2y^{\text{'}}-4y\\ \Rightarrow y^{"}-4y^{\text{'}}+4y=0\end{array}$

This is the required differential equation of the given curve.

16. Let the given statement as

P(n)= $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … + $\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$

If n=1, then

P(1)= $\frac{1}{1.4}$ = $\frac{1}{4}$ = $\frac{1}{(3.1+1)}$ = $\frac{1}{4}$

which is true.

Consider P(k)be true for some positive integer k

P(k)= $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … + $\frac{1}{(3k-2)(3k+1)}$ = $\frac{k}{(3k+1)}$ ------------------(1)

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

P(k+1)= $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}$ + … + $\frac{1}{(3k-2)(3k+1)}+\frac{1}{[3(k+1)-2][3(k+1)+1]}$

By using (1),

$\frac{k}{(3k+1)}+\frac{1}{(3k+3-2)(3k+3+1)}$

= $\frac{k}{(3k+1)}+\frac{1}{(3k+1)(3k+4)}$

= $\frac{1}{(3k+1)}\left\{k+\frac{1}{3k+4}\right\}$

= $\frac{1}{(3k+1)}\left\{\frac{k(3k+4)+1}{3k+4}\right\}$

= $\frac{1}{(3k+1)}\left\{\frac{3k^2+4k+1}{3k+4}\right\}$

= $\frac{1}{(3k+1)}\left\{\frac{3k^2+3k+k+1}{3k+4}\right\}$

= $\frac{1}{(3k+1)}\left\{\frac{3k(k+1)+(k+1)}{3k+4}\right\}$

= $\frac{1}{(3k+1)}\frac{(3k+1)(k+1)}{3k+4}$

= $\frac{k+1}{3k+4}=\frac{k+1}{3(k+1)+1}$

? P(k+1) is true whenever P(k) is true.

Therefore, from the principle of mathematical induction, the P(n) is true for all natural number n.

Given: Equation of the family of curves  $y=a{e}^{3x}+b{e}^{-2x}..........(i)$

Differentiating both sides with respect to x, we get:

$y^{\text{'}}=3a{e}^{3x}-2b{e}^{-2x}..........(ii)$

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

$y^{"}=9a{e}^{3x}-4b{e}^{-2x}..........(iii)$

Multiplying equation (i) with (ii) and then adding it to equation (ii), we get:

$\begin{array}{l}\left(2a{e}^{3x}+2b{e}^{-2x}\right)+\left(3a{e}^{3x}-2b{e}^{-2x}\right)=2y+y^{\text{'}}\\ \Rightarrow 5a{e}^{3x}=2y+y^{\text{'}}\\ \Rightarrow a{e}^{3x}=\frac{2y+y^{\text{'}}}{5}\end{array}$

Now, multiplying equation (i) with (iii) and subtracting equation (ii) from it, we get:

$\begin{array}{l}\left(3a{e}^{3x}+2b{e}^{-2x}\right)-\left(3a{e}^{3x}-2b{e}^{-2x}\right)=3y-y^{\text{'}}\\ \Rightarrow 5b{e}^{-2x}=3y-y^{\text{'}}\\ \Rightarrow b{e}^{-2x}=\frac{3y-y^{\text{'}}}{5}\end{array}$

Substituting the values of $a{e}^{3x}$ and $b{e}^{-2x}$ in equation (iii), we get:

$\begin{array}{l}{y}^{"}=9.\frac{\left(2y-y^{\text{'}}\right)}{5}+4\frac{\left(3y-y^{\text{'}}\right)}{5}\\ \Rightarrow y^{"}=\frac{18y+9y^{\text{'}}}{5}+\frac{12y-4y^{\text{'}}}{5}\\ \Rightarrow y^{"}=\frac{30y+5y^{\text{'}}}{5}\\ \Rightarrow y^{"}=6y+y^{\text{'}}\\ \Rightarrow y^{"}-y^{\text{'}}-6y=0\end{array}$

This is the required differential equation of the given curve.

Given: Equation of the family of curves  $y^2=a\left(b^2-x^2\right)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l}2y\frac{dy}{dx}=a\left(-2x\right)\\ \Rightarrow 2y{y}^{\text{'}}=-2ax\\ \Rightarrow y{y}^{\text{'}}=-ax..........(1)\end{array}$

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

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

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

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

This is the required differential equation of the given curve.

94. Kindly go through the solution

Given: Equation of the family of curves  $\frac{x}{a}+\frac{y}{b}=1..........(i)$

Differentiating both sides of the given equation with respect to x, we get:

$\begin{array}{l}\frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0\\ \Rightarrow \frac{1}{a}+\frac{1}{b}{y}^{\text{'}}=0\end{array}$

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

$\begin{array}{l}0+\frac{1}{b}{y}^{"}=0\\ \Rightarrow \frac{1}{b}{y}^{"}=0\\ \Rightarrow y^{"}=0\end{array}$

Hence, the required differential equation of the given curve is $y^{"}=0$