solve the IVP by using Laplace transform y”+ y’- 2y = 4 with y(0)= 2 and y'(0)= 1
The Correct Answer and Explanation is:
To solve the initial value problem (IVP)y′′+y′−2y=4,y(0)=2,y′(0)=1y” + y’ – 2y = 4,\quad y(0) = 2,\quad y'(0) = 1y′′+y′−2y=4,y(0)=2,y′(0)=1
using the Laplace transform, follow these steps:
Step 1: Take the Laplace transform of both sides
Using the properties of Laplace transforms:
- L{y′′(t)}=s2Y(s)−sy(0)−y′(0)\mathcal{L}\{y”(t)\} = s^2Y(s) – sy(0) – y'(0)L{y′′(t)}=s2Y(s)−sy(0)−y′(0)
- L{y′(t)}=sY(s)−y(0)\mathcal{L}\{y'(t)\} = sY(s) – y(0)L{y′(t)}=sY(s)−y(0)
- L{y(t)}=Y(s)\mathcal{L}\{y(t)\} = Y(s)L{y(t)}=Y(s)
- L{4}=4s\mathcal{L}\{4\} = \frac{4}{s}L{4}=s4
Substitute the given initial values y(0)=2y(0) = 2y(0)=2 and y′(0)=1y'(0) = 1y′(0)=1:L{y′′+y′−2y}=s2Y(s)−2s−1+sY(s)−2−2Y(s)=4s\mathcal{L}\{y” + y’ – 2y\} = s^2Y(s) – 2s – 1 + sY(s) – 2 – 2Y(s) = \frac{4}{s}L{y′′+y′−2y}=s2Y(s)−2s−1+sY(s)−2−2Y(s)=s4
Simplify:(s2+s−2)Y(s)−(2s+3)=4s(s^2 + s – 2)Y(s) – (2s + 3) = \frac{4}{s}(s2+s−2)Y(s)−(2s+3)=s4
Now solve for Y(s)Y(s)Y(s):(s2+s−2)Y(s)=4s+2s+3(s^2 + s – 2)Y(s) = \frac{4}{s} + 2s + 3(s2+s−2)Y(s)=s4+2s+3
Factor the quadratic:(s+2)(s−1)Y(s)=4s+2s+3(s + 2)(s – 1)Y(s) = \frac{4}{s} + 2s + 3(s+2)(s−1)Y(s)=s4+2s+3
Step 2: Solve for Y(s)Y(s)Y(s)
Y(s)=4s+2s+3(s+2)(s−1)Y(s) = \frac{\frac{4}{s} + 2s + 3}{(s + 2)(s – 1)}Y(s)=(s+2)(s−1)s4+2s+3
We want to simplify this expression by decomposing into partial fractions.
Let us write:Y(s)=4s(s+2)(s−1)+2s+3(s+2)(s−1)Y(s) = \frac{4}{s(s + 2)(s – 1)} + \frac{2s + 3}{(s + 2)(s – 1)}Y(s)=s(s+2)(s−1)4+(s+2)(s−1)2s+3
Let us write each part as a sum of simpler fractions:
First term:
4s(s+2)(s−1)=As+Bs+2+Cs−1\frac{4}{s(s + 2)(s – 1)} = \frac{A}{s} + \frac{B}{s + 2} + \frac{C}{s – 1}s(s+2)(s−1)4=sA+s+2B+s−1C
Multiply both sides by s(s+2)(s−1)s(s + 2)(s – 1)s(s+2)(s−1):4=A(s+2)(s−1)+B(s)(s−1)+C(s)(s+2)4 = A(s + 2)(s – 1) + B(s)(s – 1) + C(s)(s + 2)4=A(s+2)(s−1)+B(s)(s−1)+C(s)(s+2)
Choose values of sss to solve:
- Let s=0s = 0s=0:
4=A(2)(−1)⇒4=−2A⇒A=−24 = A(2)(-1) \Rightarrow 4 = -2A \Rightarrow A = -24=A(2)(−1)⇒4=−2A⇒A=−2
- Let s=−2s = -2s=−2:
4=B(−2)(−3)=6B⇒B=234 = B(-2)(-3) = 6B \Rightarrow B = \frac{2}{3}4=B(−2)(−3)=6B⇒B=32
- Let s=1s = 1s=1:
4=C(1)(3)=3C⇒C=434 = C(1)(3) = 3C \Rightarrow C = \frac{4}{3}4=C(1)(3)=3C⇒C=34
So,4s(s+2)(s−1)=−2s+23(s+2)+43(s−1)\frac{4}{s(s + 2)(s – 1)} = \frac{-2}{s} + \frac{2}{3(s + 2)} + \frac{4}{3(s – 1)}s(s+2)(s−1)4=s−2+3(s+2)2+3(s−1)4
Second term:
2s+3(s+2)(s−1)=Ds+2+Es−1\frac{2s + 3}{(s + 2)(s – 1)} = \frac{D}{s + 2} + \frac{E}{s – 1}(s+2)(s−1)2s+3=s+2D+s−1E
Multiply both sides:2s+3=D(s−1)+E(s+2)2s + 3 = D(s – 1) + E(s + 2)2s+3=D(s−1)+E(s+2)
Expand and match coefficients:2s+3=Ds−D+Es+2E=(D+E)s+(−D+2E)2s + 3 = Ds – D + Es + 2E = (D + E)s + (-D + 2E)2s+3=Ds−D+Es+2E=(D+E)s+(−D+2E)
Match:
- D+E=2D + E = 2D+E=2
- −D+2E=3-D + 2E = 3−D+2E=3
Solve the system:
From the first: D=2−ED = 2 – ED=2−E
Substitute into second:−(2−E)+2E=3⇒−2+E+2E=3⇒3E=5⇒E=53,D=13-(2 – E) + 2E = 3 \Rightarrow -2 + E + 2E = 3 \Rightarrow 3E = 5 \Rightarrow E = \frac{5}{3},\quad D = \frac{1}{3}−(2−E)+2E=3⇒−2+E+2E=3⇒3E=5⇒E=35,D=31
So,2s+3(s+2)(s−1)=13(s+2)+53(s−1)\frac{2s + 3}{(s + 2)(s – 1)} = \frac{1}{3(s + 2)} + \frac{5}{3(s – 1)}(s+2)(s−1)2s+3=3(s+2)1+3(s−1)5
Step 3: Combine all partial fractions
Y(s)=−2s+(23+13)1s+2+(43+53)1s−1Y(s) = \frac{-2}{s} + \left(\frac{2}{3} + \frac{1}{3}\right)\frac{1}{s + 2} + \left(\frac{4}{3} + \frac{5}{3}\right)\frac{1}{s – 1}Y(s)=s−2+(32+31)s+21+(34+35)s−11
Simplify:Y(s)=−2s+1s+2+3s−1Y(s) = \frac{-2}{s} + \frac{1}{s + 2} + \frac{3}{s – 1}Y(s)=s−2+s+21+s−13
Step 4: Inverse Laplace Transform
Now take the inverse Laplace transform:
- L−1{1s}=1\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1L−1{s1}=1
- L−1{1s+a}=e−at\mathcal{L}^{-1}\left\{\frac{1}{s + a}\right\} = e^{-at}L−1{s+a1}=e−at
- L−1{1s−a}=eat\mathcal{L}^{-1}\left\{\frac{1}{s – a}\right\} = e^{at}L−1{s−a1}=eat
So:y(t)=−2+e−2t+3ety(t) = -2 + e^{-2t} + 3e^{t}y(t)=−2+e−2t+3et
Final Answer:
y(t)=−2+e−2t+3et\boxed{y(t) = -2 + e^{-2t} + 3e^{t}}y(t)=−2+e−2t+3et
Explanation
This problem involves solving a second-order linear differential equation with constant coefficients and a constant nonhomogeneous term using Laplace transforms. The key advantage of using Laplace transforms is that it converts differential equations into algebraic equations, which are easier to solve.
First, we applied the Laplace transform to each term in the differential equation. Derivatives of y(t)y(t)y(t) transform into expressions involving Y(s)Y(s)Y(s), the Laplace transform of y(t)y(t)y(t), and initial conditions y(0)y(0)y(0) and y′(0)y'(0)y′(0). These initial values are substituted immediately to simplify the transformed equation.
Next, we solved for Y(s)Y(s)Y(s), which is the Laplace transform of the unknown function y(t)y(t)y(t). This required rearranging the algebraic expression and then breaking it into simpler components using partial fraction decomposition. The expression was split into two parts: one rational expression involving all three factors s(s+2)(s−1)s(s + 2)(s – 1)s(s+2)(s−1), and another rational expression involving only (s+2)(s−1)(s + 2)(s – 1)(s+2)(s−1).
Each fraction was decomposed into simpler terms using the method of partial fractions. This process involves matching coefficients to determine unknown constants. Once all constants were found, we added like terms for clarity.
Finally, we used known inverse Laplace transforms to return to the time domain. The result is a linear combination of exponential functions and a constant, giving the solution to the original differential equation. This method ensures accuracy and is especially effective when initial conditions are provided.
