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Question Number 116052 by Study last updated on 30/Sep/20

$$\frac{d^{2}y}{{dx}^2}+25y=0y=?$$

Commented by mohammad17 last updated on 30/Sep/20

$$(D^2+25)y=0$$$$$$$$(m^2+25)=0\Rightarrow m=\pm 5i$$$$$$$$y=C_{1}cos\left(5x\right)+C_{2}sin\left(5x\right)$$$$$$$$\ll m.o\gg$$

Answered by Dwaipayan Shikari last updated on 30/Sep/20

$$\mathrm{y}={\mathrm{e}}^{\lambda \mathrm{x}}$$$${\lambda }^2{\mathrm{e}}^{\lambda \mathrm{t}}+{25\mathrm{e}}^{\lambda \mathrm{t}}=0$$$$\lambda =\pm 5\mathrm{i}$$$$\mathrm{y}={\mathrm{C}}_1\cos \left(5\right)+{\mathrm{C}}_2\sin \left(5\right)$$$$$$

Answered by Bird last updated on 30/Sep/20

$$y^{{}^{″}}+25y=0\to r^2+25=0\Rightarrow$$$$r=5iorr=-5i\Rightarrow$$$$y={ae}^{5i}+{be}^{-5i}=\alpha cos\left(5x\right)+\beta sin\left(5x\right)$$

Commented by Bird last updated on 30/Sep/20

$$sorryy={ae}^{5ix}+{be}^{-5ix}=...$$

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