9. Solving Integro-Differential Equations
An "integro-differential equation" is an equation that involves both integrals and derivatives of an unknown function.
Using the Laplace transform of integrals and derivatives, an integro-differential equation can be solved.
Similarly, it is easier with the Laplace transform method to solve simultaneous differential equations by transforming both equations and then solve the two equations in the s-domain and finally obtain the inverse to get the solution in the t-domain.
Example 1 (Integro-Differential Equation)
Solve the equation for the response i(t), given that
and i(0) = 0.
First, we find the Laplace Transform of each term in the equation:
We multiply throughout by s and use the fact that i(0) = 0 to obtain:
Solving for I and completing the square on the denominator:
`I=1/(s^2+2s+5)` `=1/((s+1)^2+4)` `=1/2 2/((s+1)^2+4)`
Taking the inverse Laplace transform to give us the current as a function of time:
Easy to understand math videos:
This is the graph of the solution we just found:
The graph of i(t).
Example 2 (Simultaneous DEs)
Solve for x(t) and y(t), given that x(0) = 4, y(0) = 3, and:
Finding the Laplace transform (using the Table of Laplace Transforms) of each differential equation yields:
First equation: `(dx)/(dt)+x+4y=10`
Take Laplace Transform:
`(s+1)X+4Y=10/s+4\ \ \ ...(1)`
Second equation: `x-(dy)/(dt)-y=0`
Take Laplace Transform:
`X-(s+1)Y=-3\ \ \ ...(2)`
We now have 2 equations in 3 unknowns (X, Y and s). We need to solve for X and Y and leave s on the right hand side of each expression. We choose to solve for Y first.
Equation `(2) × (s + 1)`:
`(s+1)X-(s+1)^2Y` `=-3(s+1)\ \ \ ...(3)`
`(1) − (3)` gives:
Solve for `Y`:
The last step above is using partial fractions.
Comparing terms gives us:
s2 terms: A + B = 3
s terms: 2A + C = 7
Constant terms: 5A = 10
So `A=2, B=1, C=3`
We can now write Y as the sum of partial fractions:
Finding the inverse Laplace of our expression for Y gives:
`y=2+e^(-t)cos\ 2t+e^(-t)sin\ 2t`
We now need to find the expression for x. Equation (2) gives us:
So `x=2+2e^(-t)cos\ 2t-` `2e^(-t)sin\ 2t`
Therefore, in summary:
`x=2(1+e^(-t)cos 2t-e^(-t)sin 2t)`
`y=2+e^(-t)cos 2t+e^(-t)sin 2t`
The rectangular plot of the solution is an interesting curve:
The graph of (x(t), y(t)), for `-3 < t < 3`, showing the point (3, 4) at `t = 0`.
The curve start at the top (at `t = -3`) and loops anticlockwise until `t=0` (at the point `(3, 4)`, and then the loop gets smaller and smaller, approaching `(2,2)` as `t->oo`.