# Use the Laplace transform to solve the given initial-value problem. y'+4y=e-4t, y(0)=5

Use the Laplace transform to solve the given initial-value problem.
${y}^{\prime }+4y=e-4t,y\left(0\right)=5$
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Step 1
To apply Laplace transform technique to obtain the solution of the given initial value problem
Step 2
Obtain an equation for $F\left(s\right)=L\left(y\left(t\right)\right)$
Recall , $L\left({y}^{\prime }\left(t\right)\right)=sL\left(y\left(t\right)\right)-f\left(0\right)$
apply Laplace tranform to
${y}^{\prime }+4y={e}^{\left(}-4t\right),y\left(0\right)=5$ , to get
$sF\left(s\right)+4F\left(s\right)-5=L\left({e}^{-4t}\right)=\frac{1}{\left(s+4\right)}$
Step 3
Find the solution y(t) by taking the inverse Laplace transform of F(s), usimg standard formula
$F\left(s\right)\left(s+4\right)=5+\frac{1}{\left(s+4\right)}$
$F\left(s\right)=\frac{5}{\left(s+4\right)}+\frac{1}{\left(s+4{\right)}^{2}}$
$⇒y\left(t\right)={L}^{-1}\left(F\left(s\right)\right)$

Step 4
ANSWER: $y\left(t\right)=5{e}^{-4t}+t{e}^{-4t}$
Check:
$y\left(t\right)=5{e}^{-4t}+t{e}^{-4t}$
${y}^{\prime }\left(t\right)=-20{e}^{-4t}+{e}^{-4t}-4t{e}^{-4t}$