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87. Given, x = 2at2 and y = at4. Differentiation w r t we get,

$\frac{dx}{dt}=4at.$ and $\frac{dy}{dt}=4a{t}^3.$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{4a{t}^3}{4at}=t^2.$

Given,  $f{x}^n$ is $y=cosx+c$

So,  $y^{|}=-sinx$

Putting the value of $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}+sinx=-sinx+sinx=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

Given,  $f{x}^n$ is $y=x^2+2x+c$

So,  $y^{|}=2x+2$

Substituting value of $y^{|}$ in the given D.E. we get,

$L.H.S.=y^{|}-2x-2=2x+2-2x-2=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

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Given $f{x}^n$ is $y=e^x+1$

Differentiating with $x$ we get,

$y^{|}=\frac{dy}{dx}=e^x$

Again,

$y^{\left|\right|}=\frac{d^{2}y}{d{x}^2}=e^x$

Substituting value of $y^{\left|\right|}$ and $y^{|}$ in the given D.E. we get

$L.H.S.=y^{\left|\right|}-y^{|}=e^x-e^x=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

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The highest order derivative present in the given D.E. is $\frac{d^{2}y}{d{x}^2}$ and its order is 2.

$\therefore$ Option (A) is correct.

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86. To prove $\frac{d}{dx}(uv·w)=\frac{du}{dx}v·w+u·\frac{dv}{dx}·w+u·v·\frac{dw}{dx}.$

By repeating application of produced rule

$=\frac{d}{dx}(uv:u)$

$=u\frac{d}{dx}(u·w)+v·w\frac{dy}{dx}$

$u\left\{v\frac{dw}{dx}+w\frac{dv}{dx}\right\}+\frac{dy}{dx}v:w.$

$=u·v·\frac{dw}{dx}+u·\frac{dv}{dx}w+\frac{dy}{dx}·v·w$

$=\frac{dy}{dx}u·w+u·\frac{dv}{dx}·w+u·v·\frac{dw}{dx}$ = R×H×S×

By togarith differentiating,

Let y = u v w

Taking log, log y = log u + log v + log w

Differentiating w r t ‘x’

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}.$

$\Rightarrow \frac{dy}{dx}=y\left[\frac{1}{4}\frac{dy}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right]$

$\Rightarrow \frac{d}{dx}(uvw)=uvw·\left[\frac{1}{u}\frac{dy}{dx}+\frac{1}{v}\frac{du}{dx}+\frac{1}{w}\frac{dur}{dx}\right]$

$=\frac{dy}{dx}v·w+u\frac{dv}{dx}·w+uv·\frac{dw}{dx}$

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

P (n)=1.2+2.3+3.4+ … +2 (n+1)= $\left[\frac{n(n+1)(n+2)}{3}\right]$

For n=1,

P (1)=1.2=2= $\frac{1(1+1)(1+2)}{3}$ = $\frac{2\times 3}{3}$ =2.

Which is true.

considerP (k) be true for some positive integer k

1.2 + 2.3 + 3.4 + … + k (k + 1) = $\left[\frac{k(k+1)(k+2)}{3}\right]$ - (1)

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

Here, 1.2 + 2.3 + 3.4 + … + k (k + 1) + (k+1) (k+2)

By using (1), we get

= $\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$

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

= $\frac{(k+1)(k+2)(k+3)}{3}$

By further simplification; $\frac{(k+1)(k+1+1)(k+1+2)}{3}$

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural no. i.e., n.

In the given D.E,

$sin\frac{dy}{dx}$ is a trigonometric function of derivative $\frac{dy}{dx}$ . So it is not a polynomial equation so its derivative is not defined.

Hence, Degree of the given D.E. is not defined.

$\therefore$ Option (D) is correct.

The highest order derivative present in the D.E. is $y^{\left|\right|}$ so its order is 2.

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

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The highest order derivative present in the D.E. is $y^{\left|\right|}$ so its order is 2.

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

The given order derivative 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.