# 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

`(di)/(dt)+2i+5int_0^ti\ dt=u(t)`

and *i*(0) = 0.

Answer

First, we find the Laplace Transform of each term in the equation:

`sI-i(0)+2I+5 I/s=1/s`

We multiply throughout by *s* and use the fact that *i*(0) = 0 to obtain:

`s^2I+2sI+5I=1`

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:

`i=1/2e^(-t)sin 2t`

### Solution Graph

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:

`(dx)/(dt)+x+4y=10`

`x-(dy)/(dt)-y=0`

Answer

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:

`sX-x(0)+X+4Y=10/s`

`sX-4+X+4Y=10/s`

`(s+1)X+4Y=10/s+4\ \ \ ...(1)`

Second equation: `x-(dy)/(dt)-y=0`

Take Laplace Transform:

`X-(sY-y(0))-Y=0`

`X-sY+3-Y=0`

`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:

`(4+(s+1)^2)Y=10/s+4+3s+3`

`=10/s+3s+7`

`=(3s^2+7s+10)/s`

Solve for `Y`:

`Y=(3s^2+7s+10)/(s(4+(s+1)^2))`

`=(3s^2+7s+10)/(s(s^2+2s+5))`

`=A/s+(Bs+c)/(s^2+2s+5)`

The last step above is using partial fractions.

`3s^2+7s+10` `=A(s^2+2s+5)+s(Bs+C)`

Comparing terms gives us:

*s*^{2} terms: *A *+* B* = 3

*s* terms: 2*A *+* C* = 7

Constant terms: 5*A* = 10

So `A=2, B=1, C=3`

We can now write *Y* as the sum of partial fractions:

`Y=2/s+(s+3)/(s^2+2s+5)`

`=2/s+(s+1)/((s+1)^2+2^2)` `+2/((s+1)^2+2^2)`

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:

`X=-3+(s+1)Y`

and since

`Y=2/s+(s+3)/(s^2+2s+5)`

`=(3s^2+7s+10)/(s(s^2+2s+5)),`

we have:

`X=-3+(s+1)((3s^2+7s+10)/(s(s^2+2s+5)))`

`=(-3s(s^2+2s+5)+(s+1)(3s^2+7s+10))/(s(s^2+2s+5))`

`=(4s^2+2s+10)/(s(s^2+2s+5))`

`=(2(s^2+2s+5)+2s^2-2s)/(s(s^2+2s+5))`

`=2/s+(2s)/((s^2+2s+5))-2/((s^2+2s+5))`

`=2/s+2(s+1)/((s+1)^2+2^2)-2 2/((s+1)^2+2^2)`

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`.