First solve the homogeneous equation:
y''' - 10y'' + 25y' = 0
The characteristic equation is
m²-10m²+25m = 0
from which the solution is:
m=0 or m=5 (multiplicity 2).
Therefore the complementary solution is:
yc=C1*e^5x + C2*x*e^5x + C3
(If you have problems up to this part, revise your school notes, or check following link under "homogeneous equations with constant coefficients":
http://en.wikipedia.org/wiki/Linear_differential_equation
or post)
Now we need to find the particular solution yp such that
yp''' -10yp''+25yp' = e^(-x) cos x + e^(5x) + x
and the solution of the given differential equation would be the sum of the complementary solution and the particular solution, namely:
y = yc + yp.
To find the particular solution by undetermined coefficients, we need to assume a solution while inserting coefficients to each term so that we can solve for the yet unknown coefficients.
If you are not already familiar with the method of undetermined coefficients, you can read the following link:
http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients
According to standard procedures, we will assume the particular solution to be the sum of the following terms:
Ae^(-x)cos(x)+Be^(-x)sin(x) (from e^(-x)cos(x) on the right hand side)
Cx²e^(5x) (from e^(5x), but since both e^(5x) and xe^(5x) are already in yc, we need to go one power higher to x²e^(5x))
Dx²+Ex (from x, but since C3 is a solution, we need to go to x²)
So the form of yp is:
yp=Ae^(-x)cos(x)+Be^(-x)sin(x)+Cx²e^(5x)+Dx²+Ex
where A,B,C,D,E are coefficients to be determined by comparison of terms.
Now substitute y in the original equation by yp, and proceed to do the differentiation and simplification to get:
e^(-x)(47Bcos(x)-23Bsin(x)-47Asin(x)-23Acos(x))) + 10Ce^(5x) +50Dx + (-20D+25E) = e^(-x) cos x + e^(5x) + x
Group terms together, and solve for A,B,C,D,E from:
(-23A+47B)e^(-x)cos(x) + (-47A-23B)e^(-x)sin(x) -10Ce^(5x) + 50D + (-20D+25E) = e^(-x) cos x + e^(5x) + x
By comparing coefficients, we get:
(-23A+47B) = 1 (coefficients of e^(-x)cos(x))
(-47A-23B) = 0 (coefficients of e^(-x)sin(x))
-10C = 1 (coefficients of e^(5x))
50D = 1 (coefficients of x)
-20D+25E = 0 (coefficients of x^0)
From the above, we can solve for the undetermined coefficicents A, B, C, D and E
A=-23/2738
B=47/2738
C=1/10
D=1/50
E=2/125
to give
yp = (-23/2738)e^(-x)cos(x) + (47/2738)e^(-x)sin(x) + (1/10)x^2e^(5x)+ x^2/50 + 2x/125
or the general solution:
y = yc + yp
= C1*e^5x + C2*x*e^5x + C3 + (-23/2738)e^(-x)cos(x) + (47/2738)e^(-x)sin(x) + (1/10)x^2e^(5x)+ x^2/50 + 2x/125
Substitute y into the original equation, differentiate and make sure you get back the right hand side after simplification.
Find the general solution using the method of undetermined coefficients.
y''' - 10y'' + 25y' = e^(-x) cos x + e^(5x) + x.
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