Question

Form the differential equation representing the family of curves $y=a{e}^{2x}+b{e}^{-2x}$ where a and b are arbitrary constants.

Answer 1

Given,

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

differentiating on both sides, we get,

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

again, differentiating on both sides, we get,

$y^{\prime \prime }=4a{e}^{2x}+4b{e}^{-2x}$

$y^{\prime \prime }=4(a{e}^{2x}+b{e}^{-2x})$

$y^{\prime \prime }=4y$

$\therefore y^{\prime \prime }-4y=0$

is the required equation.