I could not figure this problem during lecture notes, please help me solve it! ANSWER ONLY THE QUESTION IN THE SQUARE. I

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I could not figure this problem during lecture notes, please help me solve it! ANSWER ONLY THE QUESTION IN THE SQUARE. I set an example on the right if it helps. Thank you.
Lecture Notes Setry J2002 6:40 AM Q1: Solve the given differential equation by undetermined coefficients-Superpos approach. Cloperpoution) y+Sy=-6e-5e Example 1: Solve the given differential equation by undetermined coefficients y-By+20y=100x¹-26xe Solution: Step1: We first solve the associated homogeneous equation y-By+20y = 0 From the auxiliary equation m²-8m+20=0 we find the roots -(-8) +√3)(20) 201 M₁ 4121 complex conjugate roots Hence the complementary function is y, = e(e, cos2x+2x) 8± √64-86 3√64-00 H Step2: Now, we assume that a particular solution is in the form of g(x) Corresponding to 100x we arrume y, Corresponding to - 26xe" we arrume The assumption for the particular solution is YY +You Ax² + x + C+ (Ex+Fe No term in this assumption duplicates a term in ye We seek to determine specific coefficients A, B, C. R. and F for which y, is a solution of sonhomogeneous DE, Sabiating y, and the derivatives 24x- +Ee+ (x+Fe=24x+8+(Ex+E+Fe Y=24+Ee+ (x+E+Fe² = 24+ (x+28+ Fje Tate the given nonhomogeneous DE, we get 24+ (x+28+)-8(24x+8+(x+E+¹) +20 (A²+BX+C+ (x+F)) ** = 100²-26xe 24+ (x+28+F)e-164x-88-8(Ex+E+Fe+204+208x+20c+20(E+) = 100-26se 24-88+20C0 -164+208=0 204 = 100 (Ex+28+P-Ex+E+F)+20(x+F)=-26x 8+√-16_8+ ****** A²+B+C (Ex+Fje Because the last equation is supposed to be an identity, the coefficients of like powers of must be equal Ans 8=4 Aus F-12/11 =-2 Thes, a particular si Sa+4+ ++(-2- " A=5-00+208=0=4 10-32+20C0 = 11/10 Sted: The general solution of the given iqti y=y, +3, y=e¹(₁2x+₁)+5² +++(-2- 12=-268-2-6E+13=0+12+110-12/1 24-88+20C=0 -164+20=0 A=5 28+F-BE+F)+20F=0 E-BE+20-26