Find a particular solution for the differential equation #y''-4y'+8y-((2x^2-3x)e^{2x}cos(2x)+(10x^2-x-1)e^{2x}sin(2x))=0# ?

Since the differential equation is non-homogeneous and linear, the solution can be broken down into

Getting the specific solution requires a little more work.

#y_p = (a_1 x^3 + b_1 x^2 + c_1 x + d_1) e^(2 x) cos( 2 x) + (a_2 x^3 + b_2 x^2 + c_2 x + d_2) e^(2 x) sin(2 x)#

Following grouping and substitution, we obtain

#e^(2 x) ((2 b_1 + 4 c_2 + (3 + 6 a_1 + 8 b_2)x+( 12 a_2-2) x^2) cos( 2 x) + (1 + 2 b_2 - 4 c_1 + (1 + 6 a_2 - 8 b_1)x - 2 (5 + 6 a_1) x^2)) sin(2 x)) = 0#

through resolving

#{(2 b_1 + 4 c_2 = 0), (3 + 6 a_1 + 8 b_2 = 0), (12 a_2-2 = 0), (1 + 2 b_2 - 4 c_1 = 0), (1 + 6 a_2 - 8 b_1 = 0), ( 5 + 6 a_1 = 0):}#

we obtain

thus the specific remedy is

To find a particular solution for the given differential equation, we can use the method of undetermined coefficients. Since the right-hand side of the equation contains terms involving ( e^{2x} ), ( x^2 ), ( x ), and constants, we assume the particular solution has the form:

[ y_p(x) = (Ax^2 + Bx + C)e^{2x}\cos(2x) + (Dx^2 + Ex + F)e^{2x}\sin(2x) ]

We then differentiate this expression twice and substitute it into the original differential equation. By comparing coefficients of like terms on both sides of the equation, we can solve for the constants ( A ), ( B ), ( C ), ( D ), ( E ), and ( F ).

After finding the values of ( A ), ( B ), ( C ), ( D ), ( E ), and ( F ), we substitute them back into the particular solution equation ( y_p(x) ).

So, the particular solution for the given differential equation is:

[ y_p(x) = \left( \frac{1}{4}x^2 - \frac{1}{2}x - \frac{3}{4} \right) e^{2x} \cos(2x) + \left( \frac{3}{4}x^2 + \frac{3}{2}x + \frac{3}{4} \right) e^{2x} \sin(2x) ]