# 8. Using Inverse Laplace Transforms to Solve Differential Equations

## Laplace Transform of Derivatives

We use the following notation:

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

### First Derivative

`ccL{g^'(t)}=ccL{(dg)/(dt)}` `=sG-g(0)`

### Second Derivative

`ccL{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:

`ccL{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:

`ccL{i''(t)} = s^2I − s\ i(0) − i\ ^'(0)`

### For the *n-*th derivative

`ccL{(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:

`ccL{(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.

#### Scientific Notebook solution

This is the way we could go about this problem using Scientific Notebook. You don't need any special software to see it (it is in HTML form.)

### Example 2

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

#### Scientific Notebook solution

### Example 3

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

#### Scientific Notebook solution

## 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*.

#### Scientific Notebook solution:

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