Skip to main content
Search IntMath

450+ Math Lessons written by Math Professors and Teachers

5 Million+ Students Helped Each Year

1200+ Articles Written by Math Educators and Enthusiasts

Simplifying and Teaching Math for Over 23 Years

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)