Laplace Transform
4.6 IVP with Piecewise Forcing Functions
Solving Initial Value Problems with Piecewise Forcing Functions
In this section, we tackle initial value problems (IVP) for second-order differential equations with constant coefficients where the forcing function [asciimath]f(t)[/asciimath] is a continuous piecewise function.
[asciimath]ay''+by'+cy=f(t)\;[/asciimath] [asciimath]y(0)=k_0, \ \ y'(0)=k_1[/asciimath]
How to Solve IVPs with Piecewise Forcing Functions using the Method of Laplace Transform
1. Write the piecewise forcing function in terms of the step function.
2. Determine the Laplace transform of the differential equation.
3. Solve the transformed equation for [asciimath]Y(s)[/asciimath].
4. Use the Laplace transform tables and the translation theorem in previous sections to determine the inverse Laplace transform.
5. If required, rewrite [asciimath]y(t)[/asciimath] in piecewise format.
Solve the given initial value problem.
[asciimath]y''-3y'-10y[/asciimath] [asciimath]=5-3tu_2(t),[/asciimath] [asciimath]y(0)=0,[/asciimath] [asciimath]y'(0)=4[/asciimath]
Show/Hide Solution
1. The forcing function [asciimath]f(t)[/asciimath] is already in the step-modulated form with [asciimath]u_2(t)=u(t-2)[/asciimath].
2. Taking the Laplace transform of the equation yields
[asciimath]\mathcal{L}{y''}+ \mathcal{L}{-3y''} +[/asciimath] [asciimath]\mathcal{L}{-10y}[/asciimath] [asciimath]=\mathcal{L}{5}+ \mathcal{L}{-3tu(t-2)}[/asciimath]
Letting [asciimath]Y(s)=\mathcal{L}{y}[/asciimath] and recognizing that [asciimath]\mathcal{L}{tu(t-2)}=e^(-2s)\mathcal{L}{t+2}[/asciimath] (Applying Equation 4.5.3), we obtain
[asciimath]s^2Y(s)-sy(0)-y'(0)-3(sY(s)-y(0))-10Y(s)=5/s-3e^(-2s)(1/s^2+2/s)[/asciimath]
Applying the initial conditions, we get
[asciimath]s^2Y(s)-4-3sY(s)-10Y(s)=5/s-3e^(-2s)(1/s^2+2/s)[/asciimath]
3. Solving for [asciimath]Y(s)[/asciimath] yields
[asciimath](s^2-3s-10)Y(s)=5/s-3e^(-2s)(1/s^2+2/s)+4[/asciimath]
[asciimath]Y(s)=5/(s(s^2-3s-10) )-(3e^(-2s))/(s^2(s^2-3s-10) ) -(6e^(-2s))/(s(s^2-3s-10) )+4/(s^2-3s-10)[/asciimath]
[asciimath]Y(s)=1/(s^2-3s-10 )(4)+ 1/(s(s^2-3s-10) )(5-6e^(-2s)) +[/asciimath] [asciimath]1/(s^2(s^2-3s-10) )(-3e^(-2s) )[/asciimath]
Factoring the denominators yields
[asciimath]Y(s)=1/((s+2)(s-5) )(4) +1/(s(s+2)(s-5) )(5-6e^(-2s))+[/asciimath] [asciimath] 1/(s^2(s+2)(s-5) )(-3e^(-2s))[/asciimath]
4. To find [asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath], we note that
[asciimath]Y(s)=4F(s)+(5-6e^(-2s))G(s)+(-3e^(-2s)) H(s)[/asciimath]
where
[asciimath]F(s)=1/((s+2)(s-5) )[/asciimath] [asciimath]=-1/7(1/(s+2))+1/7(1/(s-5))[/asciimath]
[asciimath]G(s)=1/(s(s+2)(s-5) )[/asciimath] [asciimath]=-1/10(1/s)+1/14(1/(s+2))+1/35(1/(s-5))[/asciimath]
[asciimath]H(s)=1/(s^2(s+2)(s-5) )[/asciimath]
[asciimath]=3/100(1/s)-1/10(1/s^2) -1/28(1/(s+2))+1/175(1/(s-5))[/asciimath]
Computing the inverse Laplace transform of [asciimath]F(s)[/asciimath], [asciimath]G(s)[/asciimath] and [asciimath]H(s)[/asciimath], we obtain
[asciimath]f(t)=\mathcal{L}^-1{F(s)}[/asciimath] [asciimath]=-1/7e^(-2t)+1/7e^(5t)[/asciimath]
[asciimath]g(t)=\mathcal{L}^-1{G(s)}[/asciimath] [asciimath]=-1/10+1/14e^(-2t)+1/35e^(5t)[/asciimath]
[asciimath]h(t)=\mathcal{L}^-1{H(s)}[/asciimath] [asciimath]=3/100-(t)/10-1/28e^(-2t)+1/175e^(5t)[/asciimath]
To make the inverse process easier, let’s rewrite [asciimath]Y(s)[/asciimath] first.
[asciimath]Y(s)=4F(s)+5G(s)-3e^(-2s)(2G(s)+H(s))[/asciimath]
Taking the inverse transform and using the translation theorem for the terms with the exponential term, we obtain
[asciimath]y(t)=\mathcal{L}^-1{Y(s)} =[/asciimath] [asciimath]4\mathcal{L}^-1{F(s)} +5\mathcal{L}^-1{G(s)}[/asciimath] [asciimath]-3\mathcal{L}^-1{e^(-2s)(2G(s)+H(s)) }[/asciimath]
[asciimath]=4f(t)+5g(t)-3u(t-2)(2g(t-2)+h(t-2))[/asciimath]
[asciimath]=4(-1/7e^(-2t)+1/7e^(5t))[/asciimath][asciimath]+5(-1/10+1/14e^(-2t)+1/35e^(5t))-[/asciimath] [asciimath]3u_2(t)[2(-1/10+1/14e^(-2(t-2))+1/35e^(5(t-2)))+(3/100-(t-2)/10-1/28e^(-2(t-2))+1/175e^(5(t-2)))][/asciimath]
Try an Example
The current [asciimath]I[/asciimath] in an [asciimath]LC[/asciimath] series circuit is governed by the following initial value problem. Determine the current in terms of [asciimath]t[/asciimath].
[asciimath]I''(t)+9I(t)[/asciimath] [asciimath]={(1 \ \ \ 0lttlt1),(-1 \ \ \ 1lttlt2) ,(0 \ \ \ \ \ \ \ \ \ 2ltt):}[/asciimath] [asciimath]I(0)=0,[/asciimath] [asciimath]I'(0)=0[/asciimath]
Show/Hide Solution
1. The forcing function [asciimath]f(t)[/asciimath] can be written in terms of the step function as
[asciimath]f(t)= 1 + u(t-1)(-1-1)[/asciimath] [asciimath]+u(t-2)(0-(-1))[/asciimath]
[asciimath]=1-2u(t-1)+u(t-2)[/asciimath]
2. Taking the Laplace transform of the equation yields
[asciimath]\mathcal{L}{I''}+[/asciimath] [asciimath]\mathcal{L}{9I}[/asciimath] [asciimath]=\mathcal{L}{1-2u(t-1)+u(t-2)}[/asciimath]
Letting [asciimath]J(s)=\mathcal{L}{I}[/asciimath] , we obtain
[asciimath]s^2J(s)+9J(s)=1/s-(2e^-s)/s+e^(-2s)/s[/asciimath]
3. Solving for [asciimath]J(s)[/asciimath] yields
[asciimath]J(s)=1/(s(s^2+9))-(2e^-s)/(s(s^2+9))+e^(-2s)/(s(s^2+9))[/asciimath]
4. To find [asciimath]I(t)=\mathcal{L}^-1{J(s)}[/asciimath], we note that
[asciimath]J(s)=G(s)-2e^-sG(s)+e^(-2s)G(s)[/asciimath]
where
[asciimath]G(s)=1/(s(s^2+9))[/asciimath] [asciimath]=1/9(1/s)-1/9(s/(s^2+9))[/asciimath]
Computing the inverse Laplace transform of [asciimath]G(s)[/asciimath] yields
[asciimath]g(t)=\mathcal{L}^-1{G(s)}[/asciimath] [asciimath]=1/9-1/9cos(3t)[/asciimath]
Using the translation theorem, we obtain
[asciimath]I(t)=\mathcal{L}^-1{J(s)}[/asciimath] [asciimath]=\mathcal{L}^-1{G(s)}[/asciimath] [asciimath]-2\mathcal{L}^-1{e^-sG(s)}[/asciimath] [asciimath]+\mathcal{L}^-1{e^(-2s)G(s)}[/asciimath]
[asciimath]=g(t)-2g(t-1)u(t-1)+g(t-2)u(t-2)[/asciimath]
[asciimath]=1/9(1-cos(3t))[/asciimath] [asciimath]-2/9(1-cos(3(t-1)))u(t-1)[/asciimath] [asciimath]+1/9(1-cos(3(t-2)))u(t-2)[/asciimath]
5. This can be written as the piecewise function
[asciimath]I(t)[/asciimath] [asciimath]=-1/9{(cos(3t)-1\ quad\ \ \ 0lttlt1),(1+cos(3t)-2cos(3t-3) \ \ \ 1lttlt2) ,(cos(3t)-2cos(3t-3)+cos(3t-6) \ \ tgt2):}[/asciimath]
The figure below depicts the graph of the current [asciimath]I(t)[/asciimath].
Try an Example
Section 4.6 Exercises
- Solve the following initial value problem. Only provide the solution for [asciimath]2 le t lt 3[/asciimath].
[asciimath]y'' +10 y' + 26 y = {(3, 2 le t lt 3),(0,t ge 3 or t lt 2):}\ ,[/asciimath] [asciimath]y(0) = y'(0) = 0[/asciimath]
Show/Hide Answer
[asciimath]y(t)=3/26(1-e^(-5(t-2))cos(t-2)-5 e^(-5(t-2))sin(t-2))[/asciimath]
- The solution to the IVP
[asciimath]y'' - 5 y' + 6 y = {(1, 0 le t lt 6),(0,t ge 6):}, \quad\ y(0) = y'(0) = 0[/asciimath]
is in the form [asciimath]y(t) = f(t) - g(t) u_6(t)[/asciimath]. Find functions [asciimath]f(t)[/asciimath] and [asciimath]g(t)[/asciimath].
Show/Hide Answer
[asciimath]f(t)=1/6+1/3 e^(3t)-1/2 e^(2t)[/asciimath]
[asciimath]g(t)=1/6+1/3 e^(3(t-6))-1/2 e^(2(t-6))[/asciimath]
- The solution to the IVP
[asciimath]y'' - 3 y' + 2 y = {(1, 0 le t lt 9),(0,t ge 9):}, \quad \ y(0) = y'(0) = 0[/asciimath]
is in the form [asciimath]y(t) = f(t) - g(t) u_9(t)[/asciimath]. Find functions [asciimath]f(t)[/asciimath] and [asciimath]g(t)[/asciimath].
Show/Hide Answer
[asciimath]f(t)=1/2+1/2 e^(2t)-e^(t)[/asciimath]
[asciimath]g(t)=1/2+1/2 e^(2(t-9))-e^((t-9))[/asciimath]
