 # d^2y/dx^2−2dy/dx+10y=0 where x=0;y=0 and dy/dx=4?x=0;y=0 and dy/dx=4? Joni Kenny 2021-06-25 Answered

${d}^{2}\frac{y}{{dx}^{2}}-2\frac{dy}{dx}+10y=0$ where $x=0;y=0$ and $\frac{dy}{dx}=4?x=0;y=0$ and $\frac{dy}{dx}=4$?

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The given differential is ${d}^{2}\frac{y}{{dx}^{2}}-2\frac{dy}{dx}+10y=0$ where $x=0;y=0$ and $\frac{dy}{dx}=4$.
The auxilary equation is given by ${m}^{2}-2m+10=0$
$\to m=-2+\left(-\sqrt{4-40}\right)\to m=-2±2\left(\sqrt{1-10}\right)\to m=-1±\sqrt{1-10}\to m=-1±\sqrt{9}\to m=-1±3i$
Therefore the general olution is given by $y\left(x\right)={e}^{-x}\left({c}_{1}\mathrm{cos}\left(3x\right)+{c}_{2}\mathrm{sin}\left(3x\right)\right)$
When $x=0,y=0$ therefore ${c}_{1}=0$. Therefore $y={c}_{2}{e}^{-x}\left(\mathrm{sin}\left(3x\right)\right)\to dy/x={c}_{2}{e}^{-x}\left(-\mathrm{sin}\left(3x\right)+3\mathrm{cos}\left(3x\right)\right)$
Again when $x=0,\frac{dy}{dx}=4$, this implies that $4=3e2$
Therefore the solution is $y=\left(\frac{4{e}^{-x}}{3}\right)\mathrm{sin}\left(3x\right)$