I attempted to solve the following second-order differential equation using the Laplace transform method:
$$2y'' + 3y' + y = g(t)$$
with the initial conditions $y(0) = y'(0) = 0$.
The input function $g(t)$ is defined as follows:
$$g(t)=\begin{cases} t, & 4<t<5 \\ 0,& \text{otherwise}\end{cases}$$
First, I applied the Laplace transform to both sides of the equation to obtain the transformed equation in the s-domain. Then, I rearranged the equation to solve for $Y(s)$, the Laplace transform of $y(t)$.
Next, I applied partial fraction decomposition to express the transformed equation in a simpler form. I found the inverse Laplace transform of each term using tables and formulas. However, the resulting solution did not match the expected value of $y(4.5) = 0.20438$.
I'm looking for guidance and suggestions on how to correctly solve this differential equation and find the accurate value of $y(4.5)$. If there are any alternative methods or insights that could be helpful, please share them with me.
Thank you for your assistance and expertise in advance!
Steps:
Step 1: Define the differential equation:
$$2y" + 3y' + y = g(t)$$
Step 2: Define the input function $g(t)$:
$$g(t)=\begin{cases} t, & 4<t<5 \\ 0,& \text{otherwise}\end{cases}$$
Step 3: Apply the Laplace transform to both sides of the differential equation. Use the following formulas:
$$L{y"} = s^2Y(s) – sy(0) – y'(0)$$
$$L{y'} = sY(s) – y(0)$$
$$L{y} = Y(s)$$
After applying the Laplace transform, the differential equation becomes:
$$2(s^2Y(s) – sy(0) – y'(0)) + 3(sY(s) – y(0)) + Y(s) = G(s)$$
Step 4: Apply the initial conditions:
$y(0) = 0$
$y'(0) = 0$
Substitute these initial conditions into the Laplace transformed equation.
Step 5: Substitute the Laplace transform of the input function $g(t)$ into the equation. Use the following formula for the Laplace transform of $g(t)$:
$$L(g(t)) = G(s) = 1/s^2 – (e^{-4s}/s^2) + (e^{-5s}/s^2)$$
For each interval of $g(t)$, substitute the corresponding Laplace transform into the equation.
$$Y(s)[2s^2 + 3s + 1] = 1/s^2 – (e^{-4s}/s^2) + (e^{-5s}/s^2)$$
$$Y(s) =\frac{1/s^2 – (e^{-4s}/s^2) + (e^{-5s}/s^2)}{2s^2 + 3s + 1}$$
Stuck at this point!
Best Answer
First of all your Laplace transform of $g$ si wrong. Since $$g(t)=t(u(t-4)-u(t-5))=(t-4)u(t-4)+4u(t-4)-(t-5)u(t-5)-5u(t-5)$$ then $$\mathcal{L}(g)(s)=\frac{e^{-4s}}{s^2}+\frac{4e^{-4s}}{s} -\frac{e^{-5s}}{s^2}-\frac{5e^{-5s}}{s} =\frac{e^{-4s}(4s+1)}{s^2}-\frac{e^{-5s}(5s+1)}{s^2}.$$ The next step is to compute the inverse Laplace transform of $$Y(s)=\frac{e^{-4s}(4s+1)-e^{-5s}(5s+1)}{s^2(2s^2 + 3s + 1)} =e^{-4s}\frac{4s+1}{s^2(2 s + 1) (s + 1)}-e^{-5s}\frac{5s+1}{s^2(2 s + 1) (s + 1)} .$$ Find the inverse transform of the rational functions $$\frac{4s+1}{s^2(2 s + 1) (s + 1)}\quad \text{and}\quad \frac{5s+1}{s^2(2 s + 1) (s + 1)}$$ using Partial Fraction Decomposition and then apply time shifting.