Solving initial value problems with the Laplace transform
Let
Since 
we have
we have
clearvars
syms s t Y % define symbolic variables s,t,Y
Example 1: 
Take the Laplace transform of the right-hand side function 
:
:f = exp(2*t);
F = laplace(f)
Find 
:
:Y1 = s*Y-4;
Solve the equation 
for 
:
for :Sol = solve(Y1-3*Y==F,Y)
Find the partial fraction decomposition by hand: 
, multiply by 
:
, multiply by :
Plugging in 
gives 
or 
.
gives or .Plugging in 
gives 
or 
.
gives or .Find the partial fraction decomposition using Matlab:
partfrac(Sol)
Now we can apply the inverse Laplace transform: Using 
we obtain
we obtain
Use ilaplace to find the inverse Laplace transform:
sol = ilaplace(Sol)
Example 2: 
Here 
and 
has a double root 
.
and has a double root .f = exp(-t);
F = laplace(f)
Find :
:Y1 = s*Y-4;
Find 
:
:Y2 = s*Y1-5;
Solve the equation 
for :
for :Sol = solve(Y2+3*Y1+2*Y==F,Y)
Find the partial fraction decomposition by hand:
We first need the roots of the characteristic polynomial: 
gives 
and 
, so we have 
. The denominator function for will be
gives and , so we have . The denominator function for will be
Hence we want to find a partial fraction decomposition
Multiplying by 
gives
gives
Plugging in gives 
, so 
.
gives , so .Plugging in gives 
, so 
.
gives , so .It remains to find A.
Method 1: Take the derivative of :
:
In the derivative plug in . This gives 
, so 
.
. This gives , so .Method 2: In , plug in a different value, e.g., 
this gives 
, 
, plug in a different value, e.g., this gives , Find the partial fraction decomposition using Matlab:
partfrac(Sol)
Now we can apply the inverse Laplace transform: Using 
we obtain
we obtain
Use ilaplace to find the inverse Laplace transform:
sol = ilaplace(Sol)
Example 3: 
Here and has complex roots 
.
and has complex roots .So far we just used F=laplace(f)
The full form of the command is F=laplace(f,t,s) where t is the variable for f, and s is the variable for F.
f = 3;
F = laplace(f,t,s) % since there is no explicit t in f we need to use laplace(...,t,s)
Y1 = s*Y-1;
Y2 = s*Y1-0;
Here is :
:Sol = solve(Y2-2*Y1+5*Y==F,Y)
Taking the inverse Laplace transform gives the solution 
:
:sol = ilaplace(Sol)
If we want to find this by hand we first need the partial fraction decomposition. There are two ways to do this:
Method 1: Partial fraction decomposition (real version)
partfrac(Sol)
We have 
, so 
.
, so .We have 
since the roots are 
. We want to write the term as
since the roots are . We want to write the term as
yielding
Here 
, so
, so
Method 2: Partial fraction decomposition (complex version)
partfrac(Sol,'FactorMode','full')
We get complex roots 
. If these occur once (as in this example) the partial fraction decomposition has terms
. If these occur once (as in this example) the partial fraction decomposition has terms 
with complex 
, 
.
, . After we multiply by the denominators we can find the complex number C by plugging in 
.
. Then we obtain
If the complex roots occur twice (as in Example 4 below) we will get additional terms
occur twice (as in Example 4 below) we will get additional terms
with complex 
.
.We obtain
In example 3 we have simple roots and 
, so
and , so
Example 4: 
Here and has double complex roots 
.
and has double complex roots .f = sin(t)+cos(t);
F = laplace(f)
Y1 = s*Y-1;
Y2 = s*Y1-2
We obtain
Sol = simplify( solve(Y2+Y==F,Y) )
Partial fraction decomposition by hand: (complex version)
multiply by 
:
:
plug in 
:
:
gives 
, so 
take the derivative:
plug in :
:
Partial fraction decomposition with Matlab: (complex version)
partfrac(Sol,'FactorMode','full')
sol = ilaplace(Sol)
