Question

Use the Laplace transform to solve the given initial-value problem. dy/dt-y=z,\ y(0)=0

Laplace transform
ANSWERED
asked 2021-05-16
Use the Laplace transform to solve the given initial-value problem.
\(dy/dt-y=z,\ y(0)=0\)

Answers (1)

2021-05-17
Step 1
\(\frac{dy}{dt}-y=1,\ y(0)=0\)
\(\frac{dy}{dt}-y=1\)
\(\mathcal{L}\left\{\frac{dy}{dt}-y\right\}=\mathcal{L}\{1\}\)
\(\mathcal{L}\left\{\frac{dy}{dt}\right\}-\mathcal{L}\{y\}=\frac{1}{s}\)
\(sY(s)-y(0)-Y(s)=\frac{1}{s}\)
\(sY(s)-0-Y(s)=\frac{1}{s}\)
\(sY(s)-Y(s)=\frac{1}{s}\)
\((s-1)Y(s)=\frac{1}{s}\)
\(Y(s)=\frac{1}{s(s-1)}\)
\(y(t)=\mathcal{L}^{-1}\{Y(s)\}\)
\(=\mathcal{L}^{-1}\left\{\frac{1}{s(s-1)}\right\}\)
\(=\mathcal{L}^{-1}\left\{\frac{s-(s-1)}{s(s-1)}\right\}\)
\(=\mathcal{L}^{-1}\left\{\frac{s}{s(s-1)}-\frac{s-1}{s(s-1)}\right\}\)
\(=\mathcal{L}^{-1}\left\{\frac{1}{s-1}-\frac{1}{s}\right\}\)
\(=\mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\}-\mathcal{L}^{-1}\left\{\frac{1}{s}\right\}\)
\(=e^{t}-1\)
That is
\(y(t)=e^{t}-1\)
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