5.1 Example 1
Consider the inhomogeneous linear boundary value problem (BVP) with constant coefficients
(32)
By constructing the following homotopy:
(33)
which is equivalent to
(34)
we get the functions , as it is explained in Section 4.2. It is obvious that satisfies equation (30). Moreover, from equation (31), can be found as follows:
(35)
If we put the function and in the homotopy (33) or (34), we get
(36)
and we get the following:
Hence, the solution becomes
(37)
which is the exact solution of the problem with minimum amount of calculation.
5.2 Example 2
Consider the following inhomogeneous linear BVP with variable coefficients:
(38)
In this problem, the right-hand side function is a third-order polynomial. From (15), we have
(39)
Based on the source function , if we take as
(40)
and substitute (40) into (39), we obtain
Based on these results, we have , and the homotopy is constructed in the following form:
(41)
and we get the following:
Hence, we obtain the exact solution in the following form:
with short-length calculation.
5.3 Example 3
Let us find the solution of following BVP:
(43)
As in the previous examples, using (15) we get
(44)
Based on the source function , we conclude that is in the following form:
(45)
Substituting (45) into (44), we obtain
Now, these results let us take , and construct the homotopy in the following form:
(46)
which leads to the following equations:
Hence, the exact solution becomes
(47)
which is obtained by minimum amount of calculations.
5.4 Example 4
Let us find the solution of the following first-order non-linear BVP:
(48)
Using (15) we get
(49)
Based on the source function , we find is in the following form:
(50)
Substituting (50) into (49), we obtain
Now we can take , and construct the homotopy in the following form:
(51)
which leads to the following equations:
Hence, we obtain the exact solution
(52)
with short-length calculation.