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8. Using Inverse Laplace Transforms to Solve Differential Equations

Laplace Transform of Derivatives

We use the following notation:

Later, on this page...

Subsidiary Equation


(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.

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).


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.


Taking Laplace transform of both sides gives:


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


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)`


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`.





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.`


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`



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




`=(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`

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.


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.


`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) `


`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`

Solution Graph for Example 3


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.


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)`:


`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:




So since


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)`

Solution Graph for the Application (current at time t)

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