Question

Eliminate the arbitrary constants and differentiate the following equation

$y=a{e}^{2x}+b{e}^{-2x}+c$.Is this equation correct $y^{\prime \prime \prime }=4y^{\prime }$.

If yes enter $1$ else enter $0.$

Answer 1

$y=a{e}^{2x}+b{e}^{-2x}+c$
Differentiating it w.r.t x, we get
$y^{\prime }=2a{e}^{2x}-2b{e}^{-2x}$ ...(1)
Differentiating it w.r.t x, we get
$y^{\prime \prime }=4a{e}^{2x}+4b{e}^{-2x}$
Again differentiating it w.r.t x, we get
$y^{\prime \prime \prime }=8a{e}^{2x}-8b{e}^{-2x}=4(2a{e}^{2x}-2b{e}^{-2x})$
Substituting values from (1)
$y^{\prime \prime \prime }=4y^{\prime }$