We consider the non-homogeneous problem {y''-16y} = {256x^{2}}
First we consider the homogeneous problem {y''-16y} = 0 :
1) the auxiliary equation is a r^2+br+c = =0 . ________________
2) The roots of the auxiliary equation are (enter answers as a comma separated list). ____________________ (PLEASE SHOW YOUR WORK IN DETAIL)
3) A fundamental set of solutions is (enter answers as a comma separated list). Using these we obtain the the complementary solution y_c=c_1 y_1+ c_2 y_2 for arbitrary constants c_1 and c_2 .
____________________ (PLEASE SHOW YOUR WORK IN DETAIL)
Next we seek a particular solution y_p of the non-homogeneous problem {y''-16y} = {256x^{2}} using the method of undetermined coefficients (See the link below for a help sheet)
4) Apply the method of undetermined coefficients to find y_p = ____________________ (PLEASE SHOW YOUR WORK IN DETAIL)
I have figured out the auxilirary equation r^2-16, the roots 4 and -4, and the fundamental set of solutions e^(4x),e^(-4x), but cannot find y_p. I know that A=-16 but that is as far as I can get. Please help
okay. You know that all the s^(4x) and e^(-4x) stuff will zero out. So we need some quadratic stuff that will leave a particular solution.
So, let's say that y_p is
yp = Ax^2+Bx+C
yp" = 2A
yp" - 16yp = 2A - 16(Ax^2+Bx+C)
So, we need
-16Ax^2 - 16B + 2A-16C = 256x^2
A = -16
B = 0
C = -2
y = c1*e^(4x) + c2*e^(-4x) - 16x^2 - 2
Check to be sure y"-16y = 256x^2
To find the particular solution, y_p, of the non-homogeneous problem {y''-16y} = {256x^{2}}, we can use the method of undetermined coefficients. This method involves finding a particular solution based on the form of the non-homogeneous term.
In this case, the non-homogeneous term is {256x^{2}}. Since the differential equation is a second-order differential equation, we assume that the particular solution, y_p, has the form:
y_p = Ax^2 + Bx + C
where A, B, and C are constants to be determined.
Now, let's find the derivatives of y_p:
y_p'' = 2A
y_p' = 2Ax + B
Substituting these values into the differential equation, we get:
2A - 16(Ax^2 + Bx + C) = 256x^2
Simplifying the equation, we obtain:
-16Ax^2 + (2A - 16B)x - 16C = 256x^2
Now, equating the coefficients of like terms on both sides, we have:
-16A = 256 (for the x^2 term)
2A - 16B = 0 (for the x term)
-16C = 0 (for the constant term)
Solving these equations, we find:
A = -16/256 = -1/16
B = A/8 = -(1/16)/8 = -1/128
C = 0
Therefore, the particular solution y_p is:
y_p = -(1/16)x^2 - (1/128)x
So, the general solution to the non-homogeneous problem is given by:
y = y_c + y_p = c_1e^(4x) + c_2e^(-4x) - (1/16)x^2 - (1/128)x
Therefore, the complete solution to the non-homogeneous problem is:
y = c_1e^(4x) + c_2e^(-4x) - (1/16)x^2 - (1/128)x