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Question Number 125991 by bramlexs22 last updated on 16/Dec/20
$${If}\:{a}\:{and}\:{b}\:{arbitary}\:{constants},\: \\ $$$${find}\:{a}\:{second}\:−\:{order}\:{equation} \\ $$$${which}\:{has}\:{y}\:=\:{ae}^{{x}} +{b}\:\mathrm{cos}\:{x}\:{as}\:{a} \\ $$$${general}\:{solution}. \\ $$
Answered by liberty last updated on 16/Dec/20
$$\:{by}\:{differentiating}\:{the}\:{given}\:{expression} \\ $$$${we}\:{find}\:\rightarrow{y}'\:=\:{ae}^{{x}} −{b}\:\mathrm{sin}\:{x}\:\:\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}''\:=\:{ae}^{{x}} −{b}\:\mathrm{cos}\:{x}\:\:\:\left(\mathrm{3}\right) \\ $$$${then}\:{by}\:{adding}\:{and}\:{subtracting}\:{eq}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{3}\right) \\ $$$${gives}\:\begin{cases}{{a}=\frac{{y}''+{y}}{\mathrm{2}{e}^{{x}} }}\\{{b}=\frac{{y}−{y}''}{\mathrm{2cos}\:{x}}}\end{cases} \\ $$$${substitution}\:{of}\:{these}\:{into}\:{eq}\:\left(\mathrm{2}\right)\:{we}\:{get} \\ $$$$\:\:{y}'\:=\:\frac{{y}+{y}''}{\mathrm{2}{e}^{{x}} }.{e}^{{x}} −\frac{{y}−{y}''}{\mathrm{2cos}\:{x}}.\mathrm{sin}\:{x} \\ $$$${or}\:\left(\mathrm{1}+\mathrm{tan}\:{x}\right){y}''−\mathrm{2}{y}'+\left(\mathrm{1}−\mathrm{tan}\:{x}\right){y}\:=\:\mathrm{0}. \\ $$$$ \\ $$