Question

Find the differential equation corresponding to $y={ae}^{2x}+{be}^{-3x}+{ce}^x$

Answer 1

$y=a{e}^{2x}+b{e}^{-3x}+c{e}^x..........\left(1\right)$

Differentiating above equation w.r.t. x, we have

$y^{\prime }=2a{e}^{2x}-3b{e}^{-3x}+c{e}^x..........\left(2\right)$

On subtracting ${eq}^n\left(1\right)$ from $\left(2\right)$, we have

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

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

$y\text{'}-y=a{e}^{2x}-4b{e}^{-3x}..........\left(3\right)$

Differentiating above equation w.r.t. x, we have

$y^{\prime \prime }-y^{\prime }=2a{e}^{2x}+12b{e}^{-3x}..........\left(4\right)$

Multiplying ${eq}^n\left(3\right)$ by 2, we have

$2y^{\prime }-2y=2a{e}^{2x}-8b{e}^{3x}..........\left(5\right)$

On subtracting ${eq}^n\left(5\right)$ from $\left(4\right)$, we have

$(y^{\prime \prime }-y^{\prime })-(2y^{\prime }-2y)=2a{e}^{2x}+12b{e}^{-3x}-(2a{e}^{2x}-8b{e}^{-3x})$

$y^{\prime \prime }-y^{\prime }-2y^{\prime }+2y=2a{e}^{2x}+12b{e}^{-3x}-2a{e}^{2x}+8b{e}^{-3x}$

$y^{\prime \prime }-3y\text{'}+2y=20b{e}^{-3x}..........\left(6\right)$

Differentiating above equation w.r.t. x, we have

$y\text{'}\text{'}\text{'}-3y\text{'}\text{'}+2y\text{'}=-60b{e}^{-3x}..........\left(7\right)$

Multiplying ${eq}^n\left(6\right)$ by 3, we have

$3y^{\prime \prime }-9y^{\prime }+6y=60b{e}^{-3x}..........\left(8\right)$

On adding ${eq}^n\left(7\right)\&\left(8\right)$, we have

$(y\text{'}\text{'}\text{'}-3y\text{'}\text{'}+2y\text{'})+(3y^{\prime \prime }-9y^{\prime }+6y)=-60b{e}^{-3x}+\left(60b{e}^{-3x}\right)$

$y\text{'}\text{'}\text{'}-3y\text{'}\text{'}+2y\text{'}+3y^{\prime \prime }-9y^{\prime }+6y=-60b{e}^{-3x}+60b{e}^{-3x}$

$y\text{'}\text{'}\text{'}-7y\text{'}+6y=0$

Hence, the correct answer is $y^{\prime \prime \prime }-7y^{\prime }+6y=0$.