Clear["Global`*"];
Our equation and possible substitution:
equation=(x''''[t]+5x''[t]+4x[t]==0);​​subs[m_,t_]:=Exp[mt];
Find characteristic equation and it’s roots:
charEquation=Simplify[equation/.x->(subs[m,#]&),t∈Reals]
Out[]=
4+5
2
m
+
4
m
0
In[]:=
roots=m/.Solve[charEquation,m]
Out[]=
{-,,-2,2}
Collect full solution from partial:
In[]:=
solution[t_]:=Total@Table[
k
i
subs[roots[[i]],t],{i,Length@roots}];​​solution[t]
Out[]=
-t

k
1
+
t

k
2
+
-2t

k
3
+
2t

k
4
Check:
In[]:=
Simplify[equation/.x->(solution[#]&),t∈Reals]
Out[]=
True
Get solution directly with DSolve:
In[]:=
solutionDS=x[t]/.First@DSolve[equation,x[t],t]
Out[]=

3
Cos[t]+

1
Cos[2t]+

4
Sin[t]+

2
Sin[2t]
Convert our exponential solution to trig form:
In[]:=
solutionTrig=ExpToTrig[solution[t]]
Out[]=
Cos[t]
k
1
-Sin[t]
k
1
+Cos[t]
k
2
+Sin[t]
k
2
+Cos[2t]
k
3
-Sin[2t]
k
3
+Cos[2t]
k
4
+Sin[2t]
k
4
​
Find suitable coefficients:
In[]:=
coeffs=First@Solve[ForAll[{Sin[t],Cos[t],Sin[2t],Cos[2t]},solutionTrig==solutionDS],{
k
1
,
k
2
,
k
3
,
k
4
}]
Out[]=

k
1

1
2
(

3
+

4
),
k
2

1
2
(

3
-

4
),
k
3

1
2
(

1
+

2
),
k
4

1
2
(

1
-

2
)
In[]:=
​​ComplexExpand[solutionTrig/.coeffs]==solutionDS
Out[]=
True