# Solve the following IVP for the second order linear equations y''-4y'+9y=0, y(0)=0, y'(0)=-8

Solve the following IVP for the second order linear equations
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Given:
${y}^{″}-4{y}^{\prime }+9y=0$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=-8$
The auxiliary equation of the givendifferential equation is
${m}^{2}-4m+9=0$
$m=\frac{-\left(-4\right)+-\sqrt{\left(-4{\right)}^{2}-\left(4\right)\left(1\right)\left(9\right)}}{2}$
$=\frac{4±\sqrt{16-36}}{2}$
$=\frac{\left(4±i2\sqrt{5}\right)}{2}$
$m=2±i\sqrt{5}$
Hence the general solution is
$y={e}^{2x}\left(A\mathrm{cos}\sqrt{5}x+B\mathrm{sin}\sqrt{5}x\right)$
$y\left(0\right)=0$
$⇒0=1\left(A\left(1\right)+B\left(0\right)\right)$
$⇒A=0$
$y={e}^{2x}\left(B\mathrm{sin}\sqrt{5}x\right)$
${y}^{\prime }\left(0\right)=-8$
${y}^{\prime }=2{e}^{x}\left(B\mathrm{sin}\sqrt{5}x\right)+\sqrt{5}{e}^{x}\left(B\mathrm{cos}\sqrt{5}x\right)$
$-8=0+\sqrt{5}\left(B\right)$
$B=\frac{-8}{\sqrt{5}}$
Hence, the solution is
$y=-{e}^{2x}\left(\frac{8}{\sqrt{5}}\mathrm{sin}\sqrt{5}x\right)$