y"-8y'+41y=0 y(0)=0 y'(0)=5 First , using Y for the Laplace transform of y(t), i.e., Y=Lleft{y(t)right} find the equation you get by taking the Laplace transform of the differential equation

y"-8y'+41y=0 y(0)=0 y'(0)=5 First , using Y for the Laplace transform of y(t), i.e., Y=Lleft{y(t)right} find the equation you get by taking the Laplace transform of the differential equation

Question
Laplace transform
asked 2021-02-14
\(y"-8y'+41y=0\)
\(y(0)=0\)
\(y'(0)=5\) First , using Y for the Laplace transform of y(t), i.e., \(Y=L\left\{y(t)\right\}\) find the equation you get by taking the Laplace transform of the differential equation

Answers (1)

2021-02-15
Step 1
Given
\(y"-8y'+41y=0\)
\(y(0)=0\)
\(y'(0)=5\)
Step 2
Calculation
Apply Laplace transform.
\(L\left\{y"\right\}=s^2Y(s)-sy(0)-y'(0)\)
\(L\left\{y'\right\}=sY(s)-y(0)\)
\(L\left\{y\right\}=Y(s)\)
\(L\left\{c\right\}=\frac{c}{s}\)
\(L\left\{y"\right\}-8L\left\{y'\right\}+41L\left\{y\right\}=L\left\{0\right\}\)
\(\Rightarrow s^2Y(s)-sy(0)-y'(0)-8[sY(s)-y(0)]+41Y(s)=\frac{0}{s}\)
\(\Rightarrow s^2Y(s)-0-5-8[sY(s)-0]+41Y(s)=0\)
Hence the equation is
\(s^2Y(s)-0-5-8[sY(s)-0]+41Y(s)=0\)
Now,
\(s^2Y(s)-0-5-8[sY(s)-0]+41Y(s)=0\)
\(\Rightarrow Y(s)[s^2-8s+41]=5\)
\(\Rightarrow Y(s)=\frac{5}{(s^2-8s+41)}\)
Hence the value of \(Y(s)=\frac{5}{(s^2-8s+41)}\)
Apply inverse Laplace transform
\(Y(s)=\frac{5}{(s^2-8s+41)}\)
\(=\frac{5}{(s^2-8s+16+25)}\)
\(=\frac{5}{(s-4)^2+5^2}\)
\(y(t)=L^{-1}\left(\frac{5}{(s-4)^2+5^2}\right)\)
\(\Rightarrow y(t)=e^{4t} \sin(5t)\)
Hence the value of \(y(t)=e^{4t} \sin(5t)\)
0

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