Skip to main content

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.

Answer

Taking Laplace transform of both sides gives:

`(sY-y(0))+Y=3/(s^2+9)`

`sY+Y=3/(s^2+9)` (since `y(0) = 0`)

`(s+1)Y=3/(s^2+9)`

Solving for Y and finding the partial fraction decomposition gives:

`Y=3/((s+1)(s^2+9))` `=A/(s+1)+(Bs+C)/(s^2+9)`

`3=A(s^2+9)+(s+1)(Bs+C)`

Substituting convenient values of `s` gives us:

`s=-1` gives `3=10A`, which gives `A=3/10`.

`s=0` gives `3=9A+C`, which gives `C=3/10`.

`s=1` gives `3=10A+2B+2C`, which gives us `B=-3/10`.

So

`Y=3/((s+1)(s^2+9))`

`=3/10(1/(s+1)+(-s+1)/(s^2+9))`

`=3/10(1/(s+1)-s/(s^2+9)+1/(s^2+9))`

Finding the inverse Laplace tranform gives us the solution for y as a function of t:

`y=3/10e^(-t)-3/10cos\ 3t+1/10sin\ 3t`

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.`

Answer

Taking Laplace transform of both sides and appying initial conditions of `y(0) = 1` and `y"'"(0) = 0` gives:

`{s^2Y-sy(0)-y"'"(0)}+` `2{sY-` `y(0)}+5Y=` `0`

`(s^2Y-s)+2(sY-1)+5Y=0`

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

Solving for Y and completing the square on the denominator gives:

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

`=(s+2)/((s^2+2s+1)+4)`

`=(s+2)/((s+1)^2+4)`

`=(s+1)/((s+1)^2+4)+1/2 2/((s+1)^2+4)`

Now, finding the inverse Laplace Transform gives us the solution for y as a function of t:

`y=e^(-t)cos\ 2t+1/2e^(-t) sin\ 2t`

Get the Daily Math Tweet!
IntMath on Twitter

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.

Answer

Taking Laplace transform of both sides:

`{s^2Y-sy(0)-y"'"(0)}-` `2{sY-y(0)}+Y` `=1/(s-1)`

Applying the initial condition and simplifying gives:

`(s^2Y+2s+3)-2(sY+2)+Y` `=(1)/(s-1) `

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

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


Solving for Y:

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


For the first term, we use: `Lap^{:-1:} {(n!)/((s-a)^[n+1])}=e^[at]t^n`, with a = 1 and n = 2.

So

`Lap^{:-1:}{1/2 (2)/((s-1)^3)}` `=1/2 e^t t^2 `


For the second term, we express in partial fractions:

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

`-2s+1=A(s-1)+B `


Comparing coefficients:

`-2s=As ` gives `A = -2`.

`1=-A+B ` gives `B = -1`.

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

And

`Lap^{:-1:} { - (2)/(s-1) - (1)/((s-1)^2)}` `=-2e^t - te^t`

Putting our inverse Laplace transform expressions together, the solution for y is:

`y(t)=1/2 t^2 e^t - 2e^t - te^t`

Get the Daily Math Tweet!
IntMath on Twitter

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.

Answer

We need to write the RHS of the DE in terms of unit step functions.

`(d^2i)/(dt^2)+2(di)/(dt)` `=u(t-10)-u(t-20)`

Now, taking Laplace transform of both sides gives us:

`(s^2I-s\ i(0)-i"'"(0))+2(sI-i(0))` `=(e^(-10s))/s-(e^(-20s))/s`

`s^2I+2sI` `=1/s(e^(-10s)-e^(-20s))`

`(s^2+2s)I` `=1/s(e^(-10s)-e^(-20s))`

Solving for `I` gives:

`I=1/s((e^(-10s)-e^(-20s))/(s^2+2s))` `=1/(s^2(s+2))(e^(-10s)-e^(-20s))`

We need to find the Inverse Laplace of this expression. First, we concentrate on the `1/(s^2(s+2))` part and ignore the `(e^(-10s)-e^(-20s))` part for now.

Now, we find the partial fractions: `1/(s^2(s+2))` `=A/s+B/s^2+C/(s+2)`

Multiply both sides by `s^2(s+2)`:

`1=As(s+2)+B(s+2)+Cs^2`

`s = 0` gives `1 = 2B` gives `B=1/2`

`s=-2` gives `1 = 4C` gives `C=1/4`

`s=1` gives `1 = 3A + 3B + C` gives `A= -1/4`

So `1/(s^2(s+2))` `=-1/(4s)+1/(2s^2)+1/(4(s+2))`

Now, the inverse Laplace of this expression is:

`Lap^{:-1:}{1/(s^2(s+2))}`

`Lap^{:-1:}{-1/(4s)+1/(2s^2)+1/(4(s+2))}`

`=-1/4+1/2t+1/4e^(-2t)`

So since

`I=1/(s^2(s+2))e^(-10s)-1/(s^2(s+2))e^(-20s)`,

then we have, using the Time-Displacement Theorem (see the Table of Laplace Transforms):

`i(t)=` `[-1/4*u(t-10)+` `1/2(t-10)*u(t-10)+` `{:1/4e^(-2(t-10))*u(t-10)]-` `[-1/4*u(t-20)+` `1/2(t-20)*u(t-20)+` `{:1/4e^(-2(t-20))*u(t-20)]`

`=1/4(2t-21+e^(-2(t-10)))*u(t-10)` `+1/4(41-2t-e^(-2(t-20)))*u(t-20)`

Get the Daily Math Tweet!
IntMath on Twitter

Solution Graph for the Application (current at time t)

top

Search IntMath, blog and Forum

Search IntMath

Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

The IntMath Newsletter

Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!


See the Interactive Mathematics spam guarantee.