Differential Equations

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.

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 $y^{"}=0$

In a particular solution, there are no arbitrary constant.

Hence, option (D) is correct.

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

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

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.

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 $=x{y}^{|}$

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

$=x^{2}cosx+xsinx$