# Solve the differential equation 2y"+20y'+51y=0, y(0)=2 y'(0)=0

Solve the differential equation$2y"+20{y}^{\prime }+51y=0,y\left(0\right)=2$
${y}^{\prime }\left(0\right)=0$
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Ezra Herbert
Step 1
Let $y={e}^{mt}$ be a trial solution. Then the auxiliary equation is given by $2{m}^{2}+20m+51=0as{e}^{mt}\ne 0$, for all t
Thus the solutions of auxiliary equations are
$m=-5±i\sqrt{\frac{2}{2}}$
Therefore the solution of the differential equation is given by
$y={e}^{-5}\left({c}_{1}\mathrm{cos}\left(\frac{t}{\sqrt{2}}\right)+{c}_{2}\phantom{\rule{0ex}{0ex}}sin\left(\frac{t}{\sqrt{2}}\right)\right)$
Now apply initial condition
$y\left(0\right)={e}^{-5×0}\left({c}_{1}\mathrm{cos}\left(\frac{0}{\sqrt{2}}\right)+{c}_{2}\phantom{\rule{0ex}{0ex}}sin\left(\frac{0}{\sqrt{2}}\right)\right)=1×\left(\mathrm{cos}\left(0\right){c}_{1}+\phantom{\rule{0ex}{0ex}}sin\left(0\right){c}_{2}\right)={c}_{1}$
Thus y(0)=2 gives us ${c}_{1}=2$
Step 2
Further by allpying the product rule of differention we get
${y}^{\prime }\left(x\right)=\left({e}^{-5t}\right)\left({c}_{1}\mathrm{cos}\left(\frac{t}{\sqrt{2}}\right)+{c}_{2}\phantom{\rule{0ex}{0ex}}sin\left(\frac{t}{\sqrt{2}}\right)\right)+\left({c}_{1}\left(\frac{t}{\sqrt{2}}\right)+{c}_{2}\mathrm{sin}\left(\frac{t}{\sqrt{2}}\right){\right)}^{\prime }{e}^{-5t}$
$=\left(-5{e}^{-5t}\right)\left({c}_{1}\mathrm{cos}\left(\frac{t}{\sqrt{2}}\right)+{c}_{2}\mathrm{sin}\left(\frac{t}{\sqrt{2}}\right)\right)+\left(-{c}_{1}\mathrm{sin}\left(\frac{t}{\sqrt{2}}\right)+{c}_{2}cos\left(\frac{t}{\sqrt{2}}\right)\right)\sqrt{2}{e}^{-5t}$
$=-5{e}^{-5t}\left({c}_{1}\mathrm{cos}\left(\frac{t}{\sqrt{2}}\right)+{c}_{2}\mathrm{sin}\left(\frac{t}{\sqrt{2}}\right)\right)+\left({e}^{-5t}\left(-{c}_{1}\mathrm{sin}\left(\frac{t}{\sqrt{2}}\right)+{c}_{2}\mathrm{cos}\left(\frac{t}{\sqrt{2}}\right)\right)\sqrt{2}$
Notice that
${y}^{\prime }\left(0\right)=-5{e}^{-5\ast 0}\left({c}_{1}\mathrm{cos}\left(\frac{0}{\sqrt{2}}\right)+{c}_{2}\mathrm{sin}\left(\frac{0}{\sqrt{2}}\right)\right)+\left({e}^{-5\ast 0}\left(-{c}_{1}\mathrm{sin}\left(\frac{0}{\sqrt{2}}\right)+{c}_{2}\mathrm{cos}\left(\frac{0}{\sqrt{2}}\right)\right)\sqrt{2}$
$=-5{e}^{5\ast 0}\left(\mathrm{cos}\left(0\right){c}_{1}+\mathrm{sin}\left(0\right){c}_{2}\right)+\left({e}^{-5\ast 0}\left(\mathrm{cos}\left(0\right){c}_{2}-\mathrm{sin}\left(0\right){c}_{1}\right)\right)\sqrt{2}=-5{c}_{1}+{c}_{2}\sqrt{2}$
Now y'(0)=0 gives us

$\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}{c}_{2}=10\sqrt{2}$
Therefore the required solution is given by