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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.

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94. Kindly go through the solution

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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 $y^{"}=0$

15. We can write the given statement as

P(n)=1²+3²+5²+ … + (2n – 1)²= $\frac{n(2n-1)(2n+1)}{3}$

forn=1

P(1)=1²=1= $\frac{1(2.1-1)(2.1+1)}{3}$

= $\frac{1(1)(3)}{3}=1$ which is true.

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

P(k)=1²+3²+5²+ … + (2n – 1)²= $\frac{k(2k-1)(2k+1)}{3}$ ------------------(1)

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

Here,

1²+3²+5²+ … +(2k – 1)²+(2(k+1) –1)²

By using (1),

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

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

= $\frac{(2k+1)[k(2k-1)+3(2k+1)]}{3}$

= $\frac{(2k+1)\left(2k^2-k+6k+3\right)}{3}$

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

we can write as,

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

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

= $\frac{(2k+1)(2k+3)(k+1)}{3}$

= $\frac{(2k+1)(k+1)(2k+3)}{3}$

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

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

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

93. Kindly go through the solution

In a particular solution, there are no arbitrary constant.

Hence, option (D) is correct.

14. Let the given statement be P(n) i.e.,

P(n)= $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{n}\right)=(n+1)$

If n =1

P(1)= $\left(1+\frac{1}{1}\right)$ = 2 =1+1= 2

which is true.

Assume that P(k) is true for some positive integer k i.e.,

P(k): $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{k}\right)=(k+1)$ .---------------------(1)

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

Here,

P(k+1)= $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{k}\right)+\left(1+\frac{1}{(k+1)}\right)$

By using (1), we get

(k+1). $\left(1+\frac{1}{k+1}\right)$

L.C.M.=(k+1). $\left(\frac{k+1+1}{k+1}\right)$

= (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 numbers n.

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92. Kindly go through the solution

The number of arbitrary constant is general solution of D.E of 4^th order is four.

$\therefore$ Option (D) is correct.

13. We can write given statement as

P(n): $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2n+1)}{n^2}\right)={(n+1)}^2$

If n=1, we get

P(1): $\left(1+\frac{3}{1}\right)$ =4=(1+ 1)²=2²=4

which is true.

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

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2k+1)}{k^2}\right)={(k+1)}^2$ (1)

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

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2k+1)}{k^2}\right)+\left(1+\frac{(2(k+1)+1)}{{(k+1)}^2}\right)$

By using (1)

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

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

=(k+1)²+2(k+1)+1

={(k+1)+1}²

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

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

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Kindly go through the solution