# By using Laplace transforms, solve the following differential equations subject to the given initial conditions. y"-4y'=-4te^{2t}, y_0=0, y'_0 =1 y"+2y'+10y=-6e^{-t}sin3t, y_0=0, y'_0=1

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.
$y"-4y\text{'}=-4t{e}^{2t},{y}_{0}=0,y{\text{'}}_{0}=1$

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Step 1The given differential equation is,
$y"-4{y}^{\prime }=-4t{e}^{2t}$
where ${y}_{0}=0,{y}_{0}^{\prime }=1$
Take the Laplace Transform on both sides,
$L\left({y}^{″}-4{y}^{\prime }\right)=L\left(-4t{e}^{2t}\right)$
$L\left({y}^{″}\right)-4L\left({y}^{\prime }\right)=-4L\left(t{e}^{2t}\right)$
Consider the formulae,
$L\left({y}^{″}\right)={s}^{2}L\left(y\right)-s{y}_{0}-{y}_{0}^{\prime }$
$L\left({y}^{\prime }\right)=sL\left(y\right)-{y}_{0}$
$L\left(t{e}^{2t}\right)=\frac{1}{\left(s-2{\right)}^{2}}$
Step 2
Substitute the values,
$L\left({y}^{″}\right)-4L\left({y}^{\prime }\right)=-4L\left(t{e}^{2t}\right)$
${s}^{2}L\left(y\right)-s{y}_{0}-{y}_{0}^{\prime }-4\left[sL\left(y\right)-y0\right]=-4\cdot \frac{1}{\left(s-2{\right)}^{2}}$
${s}^{2}L\left(y\right)-s\cdot \left(0\right)-\left(1\right)-4sL\left(y\right)=\frac{\left(-4\right)}{\left(s-2{\right)}^{2}}$
$L\left(y\right)\cdot \left[{s}^{2}-4s\right]=1+\frac{\left(-4\right)}{\left(s-2{\right)}^{2}}$
$L\left(y\right)=\frac{{s}^{2}-4s+4-4}{\left(s-2{\right)}^{2}\left({s}^{2}-4s\right)}$
$L\left(y\right)=\frac{1}{\left(s-2{\right)}^{2}}$
Step 3
Take the Inverse Laplace Transform both sides,
${L}^{-1}\left[L\left(y\right)\right]={L}^{-1}\left[\frac{1}{\left(s-2{\right)}^{2}}\right]$
$y={L}^{-1}\left[\frac{1}{\left(s-2{\right)}^{2}}\right]$
$y=t{e}^{2t}$
Thus, the solution of the given integral is $y=t{e}^{2t}$ii)
The given differential equation is,
${y}^{″}+24{y}^{\prime }+10y=-6{e}^{-t}\mathrm{sin}3t$

Step 4
Take the Laplace Transform on both sides,
$L\left({y}^{″}+24{y}^{\prime }+10y\right)=L\left[-6{e}^{-t}\mathrm{sin}\left(3t\right)\right]$
$L\left({y}^{″}\right)+24L\left({y}^{\prime }\right)+10L\left(y\right)=-6L\left[{e}^{-t}\mathrm{sin}\left(3t\right)\right]$
Use the same formulae as mentioned above,
${s}^{2}L\left(y\right)-0-1+2\left[sL\left(y\right)-0\right]+10L\left(y\right)=\left(-6\right)\cdot \frac{3}{\left(s+1{\right)}^{2}+9}$
${s}^{2}L\left(y\right)-1+2sL\left(y\right)+10L\left(y\right)=\frac{-18}{{s}^{2}+2s+1+9}$
$\left[{s}^{2}+2s+10\right]L\left(y\right)=1-\frac{18}{{s}^{2}+2s+10}$
$L\left(y\right)=\frac{{s}^{2}+2s+10-18}{\left({s}^{2}+2s+10{\right)}^{2}}$
$L\left(y\right)=\frac{{s}^{2}+2s-8}{\left({s}^{2}+2s+10{\right)}^{2}}$

Step 5

Take the Inverse Laplace Transform both sides,
${L}^{-1}\left[L\left(y\right)\right]={L}^{-1}\left[\frac{{s}^{2}+2s-8}{\left({s}^{2}+2s+10{\right)}^{2}}\right]$

$y=t{e}^{-t}\mathrm{cos}\left(3t\right)$

Thus, the solution of the given differential equation is $y=t{e}^{-t}\mathrm{cos}\left(3t\right)$