Find the general solution using the method of undetermined coefficients.
y''' - 10y'' + 25y' = e^(-x) cos x + e^(5x) + x.
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.
To solve the given differential equation using the method of undetermined coefficients, we assume that the particular solution can be expressed as a linear combination of functions that are similar to the right-hand side of the equation.
The equation is a third-order linear homogeneous differential equation, so we assume the particular solution has the following form:
yp(x) = Ae^(-x) cos(x) + Be^(-x) sin(x) + Ce^(5x) + Dx + E
We will now differentiate this particular solution and substitute it back into the differential equation to determine the values of the coefficients A, B, C, D, and E.
yp'(x) = (-Ae^(-x)cos(x) - Be^(-x)sin(x) + Ae^(-x)sin(x) - Be^(-x)cos(x) + 5Ce^(5x) + D)
yp''(x) = (2Ae^(-x)sin(x) -2Ae^(-x)cos(x) - 2Be^(-x)cos(x) - 2Be^(-x)sin(x) + 25Ce^(5x))
yp'''(x) = (2Ae^(-x)cos(x) + 2Ae^(-x)sin(x) - 2Be^(-x)sin(x) + 2Be^(-x)cos(x) + 125Ce^(5x))
Now, substitute these expressions into the differential equation:
yl'''(x) - 10yl''(x) + 25yl'(x) = (2Ae^(-x)cos(x) + 2Ae^(-x)sin(x) - 2Be^(-x)sin(x) + 2Be^(-x)cos(x) + 125Ce^(5x))
- 10(2Ae^(-x)sin(x) -2Ae^(-x)cos(x) - 2Be^(-x)cos(x) - 2Be^(-x)sin(x) + 25Ce^(5x))
+ 25(-Ae^(-x)cos(x) - Be^(-x)sin(x) + Ae^(-x)sin(x) - Be^(-x)cos(x) + 5Ce^(5x) + D)
Expanding and simplifying:
2Ae^(-x)cos(x) + 2Ae^(-x)sin(x) - 2Be^(-x)sin(x) + 2Be^(-x)cos(x) + 125Ce^(5x)
- 20Ae^(-x)sin(x) + 20Ae^(-x)cos(x) + 20Be^(-x)cos(x) + 20Be^(-x)sin(x) - 250Ce^(5x)
- 25Ae^(-x)cos(x) - 25Be^(-x)sin(x) + 25Ae^(-x)sin(x) - 25Be^(-x)cos(x) + 125Ce^(5x) + 25D
Simplifying further:
(2A + 20A - 25A)e^(-x)cos(x) + (2A - 2B + 20B - 25B)e^(-x)sin(x)
+ (125C - 250C + 125C)e^(5x) + 25D
To make the particular solution equal to the right-hand side of the equation, we set the coefficients of the exponential and trigonometric terms equal to the coefficients in the right-hand side of the equation:
2A + 20A - 25A = 0 -> -3A = 0 -> A = 0
2A - 2B + 20B - 25B = 1 -> -3A - 5B = 1 -> 5B = -1 -> B = -1/5
125C - 250C + 125C = 1 -> 0 = 1, which is not possible. Therefore, C = 0.
The coefficient of the linear term, 25D, should be 1. Therefore, D = 1/25.
Now, substitute these values back into the assumed particular solution:
yp(x) = 0e^(-x)cos(x) + (-1/5)e^(-x)sin(x) + 0e^(5x) + (1/25)x + E
Simplifying further:
yp(x) = (-1/5)e^(-x)sin(x) + (1/25)x + E
So, the general solution to the given differential equation is:
y(x) = yc(x) + yp(x)
where yc(x) is the general solution to the homogeneous equation and yp(x) is the particular solution we just obtained. The general solution to the homogeneous equation can be found by solving the characteristic equation associated with the homogeneous equation.
To find the general solution of the given differential equation using the method of undetermined coefficients, follow these steps:
Step 1: Find the complementary function
To find the complementary function (or the solution to the homogeneous equation), consider the characteristic equation:
r^3 - 10r^2 + 25r = 0
Factorizing, we obtain:
r(r - 5)^2 = 0
This yields three cases:
Case 1: r = 0 (multiplicity = 1)
Thus, one solution of the homogeneous equation is y1(x) = e^(0x) = 1.
Case 2: r = 5 (multiplicity = 2)
This gives us two linearly independent solutions: y2(x) = e^(5x) and y3(x) = xe^(5x).
The complementary function is given by:
y_c(x) = c1 + c2e^(5x) + c3xe^(5x)
where c1, c2, and c3 are arbitrary constants.
Step 2: Consider the particular function
The particular function (y_p(x)) is expressed in the form of the sum of all the known terms in the differential equation. Assume:
y_p(x) = A e^(-x) cos(x) + B e^(5x) + Cx + D
where A, B, C, and D are the undetermined coefficients to be determined.
Step 3: Determine the derivatives of the particular function
Calculate the first, second, and third derivatives of y_p(x) individually:
y'_p(x) = -A e^(-x) cos(x) + A e^(-x) sin(x) + 5B e^(5x) + C
y''_p(x) = 2A e^(-x) sin(x) - 2A e^(-x) cos(x) + 25B e^(5x)
y'''_p(x) = 6A e^(-x) cos(x) - 4A e^(-x) sin(x) + 125B e^(5x)
Step 4: Substitution into the differential equation and solving for coefficients
Substitute y_p(x) and its derivatives back into the original differential equation, yielding:
6A e^(-x) cos(x) - 4A e^(-x) sin(x) + 125B e^(5x) - 10(2A e^(-x) sin(x) - 2A e^(-x) cos(x) + 25B e^(5x)) + 25(-A e^(-x) cos(x) + A e^(-x) sin(x) + 5B e^(5x) + C) = e^(-x) cos(x) + e^(5x) + x
After simplification, we obtain:
(56A - 25B) e^(-x) cos(x) + (-56A - 25B) e^(-x) sin(x) + (150B + C - 10A) e^(5x) = e^(-x) cos(x) + e^(5x) + x
To satisfy this equation, we set the corresponding coefficients of each term equal to each other:
Eq 1: 56A - 25B = 1 (matching the coefficients of e^(-x) cos(x))
Eq 2: -56A - 25B = 0 (matching the coefficients of e^(-x) sin(x))
Eq 3: 150B + C - 10A = 1 (matching the coefficients of e^(5x))
Solve this system of equations for A, B, and C.
Eq 1 + Eq 2 gives:
-50B = 1
B = -1/50
Substituting B into Eq 1:
56A + 1/2 = 1
56A = 1 - 1/2
A = 1 - 1/2 / 56
A = 1 - 1/112
A = 111/112
Finally, substituting A and B into Eq 3:
150(-1/50) + C - 10(111/112) = 1
-3 + C - 10(111/112) = 1
C - 1115/112 = 1
C = 1115/112 + 1
C = 1115/112 + 112/112
C = 1227/112
Therefore, the particular solution is:
y_p(x) = (111/112)e^(-x) cos(x) - e^(5x)/50 + (1227/112)x + D
Step 5: Write the general solution
The general solution is the sum of the complementary function and the particular function:
y(x) = y_c(x) + y_p(x) = c1 + c2e^(5x) + c3xe^(5x) + (111/112)e^(-x) cos(x) - e^(5x)/50 + (1227/112)x + D
where c1, c2, c3, and D are arbitrary constants.