For complementary solution:
D2−5D+6=0⇒D2−3D−2D+6=0⇒(D−3)(D−2)=0⇒D=3,D=2
So, yc=c1e2x+c2e3x
Next, we determine particular solution by method of undetermined coefficients.
The particular solution to dx2d2y(x)−5dxdy(x)+6y(x)=xcos(2x) is of the form: yp(x)=a1cos(2x)+a2xcos(2x)+a3sin(2x)+a4xsin(2x)
Solve for the unknown constants a1,a2,a3, and a4 :
Compute dxdyp(x):dxdyp(x)=dxd(a1cos(2x)+a2xcos(2x)+a3sin(2x)+a4xsin(2x))
=−2a1sin(2x)+a2cos(2x)−2a2xsin(2x)+2a3cos(2x)+2a4xcos(2x)+a4sin(2x)
Compute dx2d2yp(x):
dx2d2yp(x)==dx2d2(a1cos(2x)+a2xcos(2x)+a3sin(2x)+a4xsin(2x))−4a1cos(2x)−4a2xcos(2x)−4a2sin(2x)−4a3sin(2x)+4a4cos(2x)−4a4xsin(2x)
Substitute the particular solution yp(x) into the differential equation:
dx2d2yp(x)−5dxdyp(x)+6yp(x)=xcos(2x)−4a1cos(2x)−4a2xcos(2x)−4a2sin(2x)−4a3sin(2x)+4a4cos(2x)−4a4xsin(2x)−5(−2a1sin(2x)+a2cos(2x)−2a2xsin(2x)+2a3cos(2x)+2a4xcos(2x)+a4sin(2x))+6(a1cos(2x)+a2xcos(2x)+a3sin(2x)+a4xsin(2x))=xcos(2x)
Simplify:
(2a1−5a2−10a3+4a4)cos(2x)+(2a2−10a4)xcos(2x)+(10a1−4a2+2a3−5a4)sin(2x)+(10a2+2a4)xsin(2x)=xcos(2x)
Equate the coefficients of cos(2x) on both sides of the equation:
2a1−5a2−10a3+4a4=0
Equate the coefficients of xcos(2x) on both sides of the equation:
2a2−10a4=1
Equate the coefficients of sin(2x) on both sides of the equation:
10a1−4a2+2a3−5a4=0
Equate the coefficients of xsin(2x) on both sides of the equation:
10a2+2a4=0
Solve the system:
a1=−1695a2=521a3=−135273a4=−525
Substitutea1,a2,a3, and a4 into yp(x)=cos(2x)a1+xcos(2x)a2+sin(2x)a3+xsin(2x)a4 :
yp(x)=−1695cos(2x)+521xcos(2x)−135273sin(2x)−525xsin(2x)
The general solution is:
Answer:
y(x)=yc(x)+yp(x)=−1695cos(2x)+521xcos(2x)−135273sin(2x)−525xsin(2x)+c1e2x+c2e3x
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