Question

For the following equation form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.​

$y=a{e}^{3x}+b{e}^{-2x}.$

Answer 1

Given family is $y=a{e}^{3x}+b{e}^{-2x}......\left(i\right)$
On differentiating w.r.t. x we get
$\frac{dy}{dx}=a\frac{d}{dx}\left(e^{3x}\right)+b\frac{d}{dx}\left(e^{-2x}\right)\Rightarrow \frac{dy}{dx}=a{e}^{3x}\frac{d}{dx}\left(3x\right)+b{e}^{-2x}\frac{d}{dx}(-2x)\Rightarrow \frac{dy}{dx}=3a{e}^{3x}-2b{e}^{-2x}...\left(ii\right)$
Again differentiating w.r.t. x, we get
$\frac{d^{1}y}{d{x}^2}=3a{e}^{3x}\frac{d}{dx}\left(3x\right)-2b{e}^{-2x}\frac{d}{dx}(-2x)=9a{e}^{3x}+4b{e}^{-2x}\Rightarrow \frac{d^{2}y}{d{x}^2}=3\left(3a{e}^{3x}\right)+4b{e}^{-2x}\Rightarrow \frac{d^{2}y}{d{x}^2}=3(\frac{dy}{dx}+2b{e}^{-2x})+4b{e}^{-2x}(\because \frac{dy}{dx}+2b{e}^{-2x}=3a{e}^{2x})\Rightarrow \frac{d^{2}y}{d{x}^2}=3\frac{dy}{dx}+6b{e}^{-2x}+4b{e}^{-2x}\Rightarrow \frac{d^{2}y}{d{x}^2}=3\frac{dy}{dx}+10b{e}^{-2x}...\left(iii\right)$
On multiplying Eq. (i) by 3 and then substracting from Eq. (ii), we get
$\frac{dy}{dx}-3y=-2b{e}^{-2x}-3b{e}^{-2x}\Rightarrow 3y-\frac{dy}{dx}=5b{e}^{-2x}\Rightarrow b=\frac{3y-\frac{dy}{dx}}{5e^{-2x}}...\left(iv\right)$
On putting the value of b in Eq. (iii), we get
$\frac{d^{2}y}{d{x}^2}=3\frac{dy}{dx}+10\frac{3y-\frac{dy}{dx}}{5e^{-2x}}.e^{-2x}\Rightarrow \frac{d^{2}y}{d{x}^2}=3\frac{dy}{dx}+2(3y-\frac{dy}{dx})\Rightarrow \frac{d^{2}y}{d{x}^2}=3\frac{dy}{dx}+6y-2\frac{dy}{dx}\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}+6y\Rightarrow \frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-6y=0$
which is the required differential equation.