Question

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

Answer 1

Step 1
To apply Laplace transform technique to obtain the solution of the given initial value problem
Step 2
Obtain an equation for \(F(s) = L(y(t))\)
Recall , \(L(y'(t))=sL(y(t))-f(0)\)
apply Laplace tranform to
\(y'+4y=e^(-4t) , y(0)=5\) , to get
\(sF(s)+4F(s)-5=L(e^{-4t})=\frac{1}{(s+4)}\)
Step 3
Find the solution y(t) by taking the inverse Laplace transform of F(s), usimg standard formula
\(F(s)(s+4)=5+\frac{1}{(s+4)}\)
\(F(s)=\frac{5}{(s+4)}+\frac{1}{(s+4)^2}\)
\(\Rightarrow y(t)=L^{-1}(F(s))\)
\(\text{So, } y(t)=5e^{-4t}+te^{-4t}\)
Step 4
ANSWER: \(y(t) = 5e^{-4t}+te^{-4t}\)
Check:
\(y(t)=5e^{-4t}+te^{-4t}\)
\(y'(t)=-20e^{-4t}+e^{-4t}-4te^{-4t}\)
\(\text{so , } y'+4y=e^{-4t} , y(0)=5\)