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Question Number 129130 by bounhome last updated on 13/Jan/21

$$solve:$$$$y^{″}+4y=cos2x$$$$$$

Answered by liberty last updated on 13/Jan/21

$$\mathrm{Homogenous}\mathrm{solution}{\mathrm{y}}_{\mathrm{h}}={\mathrm{C}}_1\cos 2\mathrm{x}+{\mathrm{C}}_2\sin 2\mathrm{x}$$$$\mathrm{W}\left({\overline{\mathrm{u}}}_1{\mathrm{u}}_2\right)=\left|\begin{array}{c}\cos 2\mathrm{x}\sin 2\mathrm{x}\\ -2\sin 2\mathrm{x}2\cos 2\mathrm{x}\end{array}\right|=2$$$${\mathrm{W}}_1=\left|\begin{array}{c}0\sin 2\mathrm{x}\\ \cos 2\mathrm{x}2\cos 2\mathrm{x}\end{array}\right|=-\frac{1}{2}\sin 4\mathrm{x}$$$${\mathrm{W}}_2=\left|\begin{array}{c}\cos 2\mathrm{x}0\\ -2\sin 2\mathrm{x}\cos 2\mathrm{x}\end{array}\right|=\cos {}^{2}2\mathrm{x}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=\frac{1}{16}\cos 4\mathrm{x}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=\frac{1}{4}\mathrm{x}+\frac{1}{16}\sin 4\mathrm{x}$$$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{16}\cos 4\mathrm{x}\cos 2\mathrm{x}+(\frac{1}{4}\mathrm{x}+\frac{1}{16}\sin 4\mathrm{x})\sin 2\mathrm{x}$$$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{16}\cos 2\mathrm{x}+\frac{1}{4}\mathrm{x}\sin 2\mathrm{x}$$$$\mathrm{General}\mathrm{solution}$$$${\mathrm{y}}_{\mathrm{G}}=({\mathrm{C}}_1+\frac{1}{16})\cos 2\mathrm{x}+({\mathrm{C}}_2+\frac{1}{4}\mathrm{x})\sin 2\mathrm{x}$$

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