8. Using Inverse Laplace Transforms to Solve Differential Equations

Laplace Transform of Derivatives

We use the following notation:

Later, on this page...

Subsidiary Equation

Application

(a) If we have the function `g(t)`, then `G(s) = G = Lap{g(t)}`.

(b) g(0) is the value of the function g(t) at t = 0.

(c) g'(0), g’’(0),... are the values of the derivatives of the function at t = 0.

If `g(t)` is continuous and g'(0), g’’(0),... are finite, then we have the following.

First Derivative

`Lap{g"'"(t)}= Lap{(dg)/(dt)}` `=sG-g(0)`

Second Derivative

`Lap{g"''"(t)}=s^2G-s\ g(0) - g"'"(0)`

We saw many of these expressions in the Table of Laplace Transforms.


NOTATION NOTE:
If instead of g(t) we have a function y of x, then Equation (2) would simply become:

`Lap{y’’(x)} = s^2Y − s\ y (0) − y’(0)`

Likewise, if we have an expression for current i and it is a function of t, then the equation would become:

`Lap{i’’(t)} = s^2I − s\ i(0) − i\ ’(0)`

For the n-th derivative

`Lap{(d^ng)/(dt^n)}` `=s^nG-s^(n-1)g(0)-s^(n-2)g ’ (0)-` `...-g^(n-1)(0)`


NOTATION NOTE: If we have y and it is a function of t, then the notation would become:

`Lap{(d^ny)/(dt^n)}` `=s^nY-s^(n-1)y(0)-s^(n-2)y ’ (0)-` `...-y^(n-1)(0)`

Subsidiary Equation

The subsidiary equation is the equation in terms of s, G and the coefficients g'(0), g’’(0),... etc., obtained by taking the transforms of all the terms in a linear differential equation.

The subsidiary equation is expressed in the form G = G(s).

Examples

Write down the subsidiary equations for the following differential equations and hence solve them.

Example 1

`(dy)/(dt)+y=sin\ 3t`, given that y = 0 when t = 0.

Solution Graph for Example 1

This is the graph of the solution we obtained in the example above.

Example 2

Solve `(d^2y)/(dt^2)+2(dy)/(dt)+5y=0`, given that `y = 1,` and `(dy)/(dt)=0,` when `t = 0.`

Solution Graph for Example 2

Here is the graph of what we just found:

Example 3

`(d^2y)/(dt^2)-2(dy)/(dt)+y=e^t`, given that y = -2, and `(dy)/(dt)=-3` when t = 0.

Solution Graph for Example 3

Application

The current i(t) in an electrical circuit is given by the DE

`(d^2i)/(dt^2)+2(di)/(dt)=0,if 0 < t < 10`

`=1,if 10 < t < 20`

`=0,if t > 20`

and i(0) = 0, i’(0) = 0.

Determine the current as a function of t.

Solution Graph for the Application (current at time t)