Ncert Solutions Maths class 12th

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 $y^{"}=0$

93. Kindly go through the solution

In a particular solution, there are no arbitrary constant.

Hence, option (D) is correct.

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.

Kindly go through the solution

91. Given, $x=a\left(cost+logtan\frac{t}{2}\right)y=asint$

Differentiating w r t we get,

$\frac{dx}{dt}=a\frac{d}{dt}\left[cost+log\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{d}{dt}\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.se{c}^2\frac{t}{2}\frac{d}{dt}\left(\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}*\frac{1}{co{s}^2\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}*\frac{1}{2}\right]$

$=a\left[-sint+\frac{1}{2sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sin2*\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sint}\right]=a\left[\frac{1-si{n}^{2}t}{sint}\right]$

$=a\frac{co{s}^{2}t}{sint}\left\{?1=co{s}^{2}x+si{n}^{2}x\right\}$

$\frac{\mathrm{bd}y}{dt}=\frac{d}{dt}\left(asint\right)=acost$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{acost}{\raisebox{1ex}{$aco{s}^{2}t$}\!\left/ \!\raisebox{-1ex}{$sint$}\right.}=\frac{sint}{cost}=tant$

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