Laplace Transform
4.4 Solving Initial Value Problems
Having explored the Laplace Transform, its inverse, and its properties, we are now equipped to solve initial value problems (IVP) for linear differential equations. Our focus will be on second-order linear differential equations with constant coefficients.
Method of Laplace Transform for IVP
General Approach:
1. Apply the Laplace Transform to each term of the differential equation. Use the properties of the Laplace Transform listed in Tables 4.1 and 4.2 to obtain an equation in terms of [asciimath]Y(s)[/asciimath]. The Laplace Transform of the derivatives are
[asciimath]\mathcal{L}{f'(t)} = sF(s) - f(0)[/asciimath]
[asciimath]\mathcal{L}{f''(t)\} = s^2F(s) - s f(0) - f'(0)[/asciimath]
2. The transforms of derivatives involve initial conditions at [asciimath]t=0[/asciimath]. Apply the initial conditions.
3. Simplify the transformed equation to isolate [asciimath]Y(s)[/asciimath].
4. If needed, use partial fraction decomposition to break down [asciimath]Y(s)[/asciimath] into simpler components.
5. Determine the inverse Laplace Transform using the tables and linearity property to find [asciimath]y(t)[/asciimath].
Shortcut Approach:
1. Find the characteristic polynomial of the differential equation [asciimath]p(s)=as^2+bs+c[/asciimath].
2. Substitute [asciimath]p(s)[/asciimath], [asciimath]F(s)=\mathcal{L}{f(t)}[/asciimath], and the initial conditions into the equation
[asciimath]Y(s)=(F(s)+a(y'(0)+sy(0))+b y(0) )/(p(s))[/asciimath] (4.4.1)
3. If needed, use partial fraction decomposition to break down [asciimath]Y(s)[/asciimath] into simpler components.
4. Determine the inverse Laplace transform of [asciimath]Y(s)[/asciimath] using the tables and linearity property to find [asciimath]y(t)[/asciimath].
Solve the initial value problem.
[asciimath]y''-5y'+6y=4e^(-2t)\ ;[/asciimath] [asciimath]y(0)=-1, \ y'(0)=2[/asciimath]
Show/Hide Solution
Using the General Approach
1. Take the Laplace Transform of both sides of the equation
[asciimath]\mathcal{L}^-1{ y''}-5\mathcal{L}^-1{ y'}+6\mathcal{L}^-1{y}=4\mathcal{L}^-1{ e^(-2t)}[/asciimath]
Letting [asciimath]Y(s)=\mathcal{L}^-1{y}[/asciimath], we get
[asciimath]s^2Y(s)-sy(0)-y'(0)-5(sY(s)-y(0))+6Y(s)=4(1/(s+2))[/asciimath]
2. Plugging in the initial conditions gives
[asciimath]s^2Y(s)+s-2-5(sY(s)+1)+6Y(s)=4(1/(s+2))[/asciimath]
3. Collecting like terms and isolating [asciimath]Y(s)[/asciimath], we get
[asciimath](s^2-5s+6)Y(s)=4/(s+2)-s+7[/asciimath]
[asciimath]Y(s)[/asciimath] [asciimath]=(4//(s+2)-s+7)/(s^2-5s+6)[/asciimath]
Multiplying both the denominator and numerator by [asciimath](s+2)[/asciimath] and factoring the denominator yields
[asciimath]Y(s)=(-s^2+5s+18)/((s+2)(s-3)(s-2))[/asciimath]
4. Using partial fraction expansion, we get
[asciimath]Y(s)=1/5 (1/(s+2))+24/5 (1/(s-3))-6 (1/(s-2))[/asciimath]
5. From Table 4.1, we see that
[asciimath]1/(s-a)[/asciimath] [asciimath]harr \ \e^(at)[/asciimath]
Taking the inverse, we obtain the solution of the equation
[asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath] [asciimath]=1/5 \ e^(-2t) +24/5 e^(3t)-6 e^(2t)[/asciimath]
Solve the initial value problem.
[asciimath]y''+4y=3sin(t) \ ;[/asciimath] [asciimath]y(0)=1, \ y'(0)=-1[/asciimath]
Show/Hide Solution
Using the Shortcut Approach
1. The characteristic polynomial is
[asciimath]p(s)=s^2+4[/asciimath]
and
[asciimath]F(s)=\mathcal{L}^-1{3sin(t)}[/asciimath] [asciimath]=3/(s^2+1)[/asciimath]
2. Substituting them together with the initial values into Equation 4.4.1, we obtain
[asciimath]Y(s)=(3//(s^2+1)+(-1+s(1)))/(s^2+4)[/asciimath] [asciimath]=(3//(s^2+1)+s-1)/(s^2+4)[/asciimath]
Multiplying both the denominator and numerator by [asciimath](s^2+1)[/asciimath] yields
[asciimath]Y(s)=(s^3-s^2+s+2)/((s^2+1)(s^2+4))[/asciimath]
3. Using partial fraction expansion, we get
[asciimath]Y(s)=1/(s^2+1)+(s-2)/(s^2+4)[/asciimath]
[asciimath]\ =1/(s^2+1)+s/(s^2+4)- 2/(s^2+4)[/asciimath]
4. From Table 4.1,
[asciimath]sin(bt)\ harr\ b/(s^2+b^2)[/asciimath] and [asciimath]cos(bt)\ harr\ s/(s^2+b^2)[/asciimath]
Taking the inverse, we obtain the solution of the equation
[asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath] [asciimath]=sin(t)+cos(2t)-sin(2t)[/asciimath]
Try an Example
Section 4.4 Exercises
- Solve the IVP by using the inverse Laplace Transform [asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath]
[asciimath]y'' +3 y' -10 y = 0, \ quad y(0) = -1, \ quad y'(0) = 2[/asciimath]
Show/Hide Answer
[asciimath]y(t)=-3/7 e^(2t)-4/7 e^(-5t)[/asciimath]
- Solve the IVP by using the inverse Laplace Transform [asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath]
[asciimath]y'' +6 y' + 13 y = 0, \ quad y(0) = 2, \ quad y'(0) = 0[/asciimath]
Show/Hide Answer
[asciimath]y(t)=e^(-3t)(2cos(2t)+3sin(2t))[/asciimath]
- Solve the IVP by using the inverse Laplace Transform [asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath]
[asciimath]y'' - 8 y' +16 y = 0, \ quad y(0) = 1, \ quad y'(0) = -1[/asciimath]
Show/Hide Answer
[asciimath]y(t)=e^(4t)(1-5t)[/asciimath]
