# 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 = ` ℒ`{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

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

### Second Derivative

ℒ`{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:

ℒ`{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:

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

### For the *n-*th derivative

ℒ`{(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:

ℒ`{(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.) This includes the graph of the solution.

### 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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