Question

D2-4D+4D=x2+ex+cos2x​

Answer 1

Answer:

(D2−4D+4)y=0

(D−2)2y=0

D=2

yh=C1e2x+C2xe2x

Solve non-homogeneous equation using Method of Variation of Parameters.

(D2−4D+4)y=x2e2xcos(2x)

y1=e2x

y2=xe2x

W(y1,y2)=y1y2′−y1′y2=(e2x)((2x+1)e2x)−(2e2x)(xe2x)=e4x

g(x)=x2e2xcos(2x)

u1=−∫y2g(x)W(y1,y2)dx=−∫xe2x⋅x2e2xcos(2x)e4xdx=−∫x3cos(2x)dx=(34x−12x3)sin(2x)+(38−34x2)cos(2x)

u2=∫y1g(x)W(y1,y2)dx=∫e2x⋅x2e2xcos(2x)e4xdx=∫x2cos(2x)dx=(12x2−14)sin(2x)+12xcos(2x)

yp=u1y1+u2y2=[(34x−12x3)sin(2x)+(38−34x2)cos(2x)]e2x+[(12x2−14)sin(2x)+12xcos(2x)]xe2x=12xe2xsin(2x)−(14x2−38)e2xcos(2x)

y=C1e2x+C2xe2x+12xe2xsin(2x)−(14x2−38)e2xcos(2x)

Answer 2

Step-by-step explanation:

I hope it's helpful for you