Solve the following differential equations using the Laplace transform frac{(dy)}{(dt)}-4y=4-2t , y(0)=0

Answers (1)

2020-12-25

Step 1
Given:
\(\frac{(dy)}{(dt)}-4y=4-2t\)
\(y(0)=0\)
\(\frac{(dy)}{(dt)}-4y=4-2t\)
Replace \(\frac{(dy)}{(dt)}\) as y'
\(y'-4y=4-2t\)
Applying Laplace transform on both sides,
\(L\left\{y'-4y\right\}=L\left\{4-2t\right\}\)
\(L\left\{y'-4y\right\}=sL\left\{y\right\}-y(0)-4L\left\{y\right\}\)
\(L\left\{4-2t\right\}=\frac{4}{s}-\frac{2}{s^2}\)
Step 2
\(sL\left\{y\right\}-y(0)-4L\left\{y\right\}=\frac{4}{s}-\frac{2}{s^2}\)
\(sL\left\{y\right\}-0-4L\left\{y\right\}=\frac{4}{s}-\frac{2}{s^2}\)
\(sL\left\{y\right\}−4L\left\{y\right\}=\frac{4}{s}-\frac{2}{s^2}\)
\(L\left\{y\right\}=\frac{4s-2}{s^2(s-4)}\)
Take the inverse Laplace transform
\(y=L^{-1}\left(\frac{4s-2}{s^2(s-4)}\right)\)
\(L^{-1}\left(\frac{4s-2}{s^2(s-4)}\right)=-\frac{7}{8}+\frac{t}{2}+\frac{7}{8}e^{4t}\)
\(y=-\frac{7}{8}+\frac{t}{2}+\frac{7}{8}e^{4t}\)

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