Solve the following differential equations using the Laplace transform and the unit step function y"+4y=g(t) y(0)=-1 y'(0)=0 , text{ where } g(t)=begin{cases}t &, tleq 25 & ,t > 2end{cases} y"-y=g(t) y(0)=1 y'(0)=2 , text{ where } g(t)=begin{cases}1 &, tleq 3t & ,t > 3end{cases}

Question
Laplace transform
asked 2021-02-02
Solve the following differential equations using the Laplace transform and the unit step function
\(y"+4y=g(t)\)
\(y(0)=-1\)
\(y'(0)=0 , \text{ where } g(t)=\begin{cases}t &, t\leq 2\\5 & ,t > 2\end{cases}\)
\(y"-y=g(t)\)
\(y(0)=1\)
\(y'(0)=2 , \text{ where } g(t)=\begin{cases}1 &, t\leq 3\\t & ,t > 3\end{cases}\)

Answers (1)

2021-02-03
Step 1 Given :The differential equations using the Laplace Transform and the unit step function
\(y"+4y=g(t)\)
\(y(0)=-1\)
\(y'(0)=0 , \text{ where } g(t)=\begin{cases}t &, t\leq 2\\5 & ,t > 2\end{cases}\)
Step 2
The given initial value problem is
\(y"+4y=g(t)\)
\(y(0)=-1\)
\(y'(0)=0 , \text{ where } g(t)=\begin{cases}t &, t\leq 2\\5 & ,t > 2\end{cases}\)
Write the step using the Heaviside step functions.
If f(t) is a function of t for \(t\geq0\) whose Laplace Transform F(s) exists , then for any constant \(a\geq0\) , the function
\(g(t)=f(t-a)u(t-a)=\begin{cases}0&\text{ if } ta \end{cases}\)
has the Laplace Transform \(e^{-as}F(s)\)
\(g(t)=(2-u(t-2))+5u(t-2)=2-u(t-2)+5u(t-2)\)
\(g(t)=2+4u(t-2)\) So, the initial value problem can be written as
\(y"+4y=2+4u(t-2) , y(0)=-1 , y'(0)=0\)
Taking the Laplace Transform of both sides , we get
\(s^2\bar y - sy(0)-y'(0)+4 \bar y =\frac{2}{s}+4e^{-2s}\)
\(s^2\bar y - s(-1)+4 \bar y =\frac{2}{s}+4e^{-2s}\)
\(s^2\bar y +s+4 \bar y =\frac{2}{s}+4e^{-2s}\)
\((s^2+4)\bar y =\frac{2}{s}+4e^{-2s}-s\)
\(\bar y =\frac{2}{s(s^2+4)}+4\frac{e^{-2s}}{s^2+4}-\frac{s}{s^2+4}\)
Taking Inverse Laplace Transform , we get
\(\text{If } L^{-1} \left[F(s)\right] = f(t) \text{ then } L^{-1} \left(F(s)\right) = \int_0^t f(t) dt\)
\(L (1) =\frac{1}{s} , L^{-1}\left(\frac{s}{s^2+4}\right) = \cos at , L^{-1}\left(\frac{1}{s^2+4}\right) = \frac{1}{a} \sin at\)
\(L^-1[\bar y] = L^{-1}\left[\frac{2}{s(s^2+4)}\right]+4L^{-1}\left[\frac{e^{-2s}}{s^2+4}\right]-L^{-1}\left[\frac {s}{s^2+4}\right]\)
\(y=\frac{2}{2} \int_0^t \sin2t dt + \frac{4e^{-2s}}{2} \sin 2t - \cos 2 t\)
\(y=-\left[\frac{\cos 2t}{2}\right]_0^t+2u(t-2)\sin2t-\cos2t\)
\(y=-\frac{1}{2}\left[\cos 2t-1\right]+2u(t-2)\sin2t-\cos2t\)
Hence, the differential equation is
\(y=-\frac{1}{2}\left[\cos 2t-1\right]+2u(t-2)\sin2t-\cos2t\)
0

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