Question

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

let

$\frac{dy}{dx}=y^{\prime }\text{and}\frac{d^{2}y}{d{x}^2}=y"$

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

differentiating both sides w.r.t. x
we get,

$\frac{dy}{dx}=\frac{d}{dx}[a{e}^{3x}+b{e}^{-2x}]$

$\frac{dy}{dx}=a{e}^{3x}\times 3+b{e}^{-2x}\times (-2)$

$y^{\prime }=3a{e}^{3x}-2b{e}^{-2x}$ (i)

Again, differentiating both sides w.r.t. x
we get,

$y"=\frac{d}{dx}[3a{e}^{3x}-2b{e}^{-2x}]$

$y"=3a{e}^{3x}\left(3\right)-2b{e}^{-2x}(-2)$

$y"=9a{e}^{3x}+4b{e}^{-2x}$ (ii)

Subtracting (ii) From (i)

$y"-y^{\prime }$

$=9a{e}^{3x}+4b{e}^{-2x}-3a{e}^{3x}+2b{e}^{-2x}$

$y"-y^{\prime }=6a{e}^{3x}+6b{e}^{-2x}$

$y^{\prime }-y^{\prime }=6(a{e}^{3x}+b{e}^{2x})$

Putting

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

$y"-y^{\prime }=6y$

$y"-y^{\prime }-6y=0$

Final Answer:
Hence, the required differential equation is

$y"-y^{\prime }-6y=0$