Maths

90. Given, x = 4t and y = $\frac{4}{t}$ Differentiating w r t. 't' we get,

$\frac{dx}{dt}=4\frac{dy}{dt}=\frac{4d\left(t^{-1}\right)}{dt}$

$=-4t^{-2}$

$\frac{-4}{t^2}$

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

Given,  $x+y=ta{n}^{-1}y$

Differentiate with 'x' we get

$\begin{array}{l}1+\frac{dy}{dx}=\frac{1}{1+y^2}\frac{dy}{dx}\\ =1+y^{|}=\frac{1}{1+y^2}{y}^{|}\\ =\left(1+y^{|}\right)\left(1+y^2\right)=y^{|}\\ =1+y^2{y}^{|}+y^{|}+y^2=y^{|}\\ =y^2{y}^{|}+y^2+1=0\end{array}$

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

11. we can write the given statement as

$P(n)=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{n(n+1)(n+2)}$ = $\frac{n(n+3)}{4(n+1)(n+2)}$

If n=1,

P(1)= $\frac{1}{1.2.3}$ = $\frac{1}{6}$ = $\frac{1(1+3)}{4(1+1)(1+2)}$ = $\frac{4}{4*2*3}$ = $\frac{1}{6}$

which is true.

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

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{k(k+1)(k+2)}$ = $\frac{k(k+3)}{4(k+1)\left(k+2\right)}$

Let us prove that P(k+1) is true,

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}$ + … + $\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}$ .

By equation (1), we get

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

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

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

= $\frac{1}{(k+1)(k+2)}\left[\frac{k\left(k^2+6k+9\right)+4}{4(k+3)}\right]$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+6k^2+9k+4}{4(k+3)}\right\}$

= $\frac{1}{(k+1)(k+2)}\left\{\frac{k^3+2k^2+k+4k^2+8k+4}{4(k+3)}\right\}$

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

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

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

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

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

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

Given $y-cosy=x$

Differentiate w.r.t 'x' we get

$\begin{array}{l}\frac{dy}{dx}-\left(-siny\right)\frac{dy}{dx}=\frac{dx}{dx}\\ \Rightarrow \frac{dy}{dx}\left[1+siny\right]=1\\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+siny}=y^{|}\end{array}$

So, L.H.S of given D.E $=\left(ysiny+cosy+x\right)y^{|}$

$\begin{array}{l}=\left(ysiny+cosy+y-cosx\right)\left[\frac{1}{1+siny}\right]\\ =\frac{y\left(1+siny\right)}{\left(1+siny\right)}=y=R.H.S\end{array}$

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

89. Given, x = sin t and y = cos2t. differentiation w r t. 't' we get,

$\begin{array}{l}\frac{dx}{dt}=cost\frac{dy}{dt}=\left(-sin2t\right)\frac{d2t}{dt}\\ -t\\ -2\left(2sintcost\right)\\ -tt\end{array}$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}$

$=\frac{-4sintcost}{cost}$

= -4 sin t

Given, $xy=logy+c$

Differentiate w.r.t. x we have

$\begin{array}{l}x\frac{dy}{dx}+y\frac{dx}{dx}=\frac{d}{dx}logy+\frac{d}{dx}C\\ \Rightarrow x\frac{dy}{dx}+y=\frac{1}{y}\frac{dy}{dx}+0\\ \Rightarrow x\frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}=-y\\ \Rightarrow \frac{dy}{dx}\left[x-\frac{1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}\left[\frac{xy-1}{y}\right]=-y\\ \Rightarrow \frac{dy}{dx}=\frac{-y^2}{xy-1}=\frac{\left(-1\right)*-y^2}{\left(-1\right)*\left(xy-1\right)}=\frac{y^2}{1-xy}\\ \therefore y^{|}=\frac{y^2}{1-xy}\end{array}$

Hence, y is a Solution of the given D.E

Given,  $y=xsinx$

So,  $y^{|}=x\frac{d}{dx}sinx+sinx\frac{dx}{dx}=xcosx+sinx$

Now, L.H.S of the given D.E $=x{y}^{|}$

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

$=x^{2}cosx+xsinx$

88. Kindly go through the solution

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

$P(n)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+?+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

For n=1,

P(1)= $\frac{1}{2.5}=\frac{1}{10}=\frac{1}{6.1+4}=\frac{1}{10}$

which is true.

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

i.e.,P(k)= $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+?+\frac{1}{(3k-1)(3k+2)}=\frac{k}{(6k+4)}$ (1)

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

Here, $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}$ + … + $\frac{1}{(3k-1)(3k+2)}+\frac{1}{(3(k+1)-1)[3(k+1)+2]}$

By using eqn.(1),

= $\frac{k}{6k+4}+\frac{1}{(3k+3-1)(3k+3+2)}$

= $\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}$

Taking 2 as common,

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

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

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

= $\frac{1}{(3k+2)}\left\{\frac{3k^2+5k+2}{\begin{array}{l}\bcancel{6k}+\bcancel{10}\\ 2(3k+5)\end{array}}\right\}$

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

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

= $\frac{(k+1)}{6k+10}$ , so we get

$\frac{(k+1)}{6(k+1)+4}$

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.

Given,  $y=Ax:$

So,  $y^{|}=\frac{Adx}{dx}=A$

Putting value of $y^{|}$ in L.H.S. of the given D.E.

L.H.S= $x{y}^{|}=xA=Ax=y$ =R.H.S

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