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
$${thanks} \\ $$
Answered by Spillover last updated on 11/Dec/25
$${Solution}\:\mathrm{11}/\mathrm{12}/\mathrm{2025} \\ $$$${y}={Ae}^{{bx}+\mathrm{1}} \\ $$$$\frac{{dy}}{{dx}}={Aebe}^{{bx}} \\ $$$$\frac{{dy}}{{dx}}={by} \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}={b} \\ $$$$\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}={b}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{1}}{{y}}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \\ $$$$ \\ $$
Answered by Spillover last updated on 11/Dec/25
$${solution}\:\mathrm{11}/\mathrm{12}/\mathrm{2025} \\ $$$$\left({b}\right){y}={A}\mathrm{sin}\:{x}+{B}\mathrm{cos}\:{x} \\ $$$$\frac{{dy}}{{dx}}={A}\mathrm{cos}\:{x}−{B}\mathrm{sin}\:{x} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−{A}\mathrm{sin}\:{x}−{B}\mathrm{cos}\:{x} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−{y} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{y}=\mathrm{0} \\ $$