Question Number 226603 by Spillover last updated on 07/Dec/25
$${Formulate}\:{the}\:{differential} \\ $$$${equation}\:{of}\:{the}\:{solution} \\ $$$$\left({a}\right){y}={Ae}^{{bx}+\mathrm{1}} \\ $$$$\left({b}\right){y}={A}\mathrm{sin}\:{x}+{B}\mathrm{cos}\:{x} \\ $$$$ \\ $$
Answered by mr W last updated on 07/Dec/25
$$\left({a}\right) \\ $$$${y}'−{by}=\mathrm{0} \\ $$$$\left({b}\right) \\ $$$${y}''+{y}=\mathrm{0} \\ $$
Commented by Spillover last updated on 07/Dec/25
$${thanks} \\ $$