Question #216156
(D-5D
1
Expert's answer
2021-07-12T15:00:55-0400
(D25D+6)y=cos3x(D^2-5D+6)y=\cos 3x

The homogeneous differential equation


(D25D+6)y=0(D^2-5D+6)y=0

The characteristic equation


r25r+6=0r^2-5r+6=0

r1=2,r2=3r_1=2,r_2=3

The general solution of the homogenepus equation is


yh=c1e2x+c2e3xy_h=c_1e^{2x}+c_2e^{3x}

Find the particular solution of the nonhomogeneous differential equation


yp=Acos3x+Bsin3xy_p=A\cos3x+B\sin3x

yp=3Asin3x+3Bcos3xy_p '=-3A\sin 3x+3B\cos3x


yp=9Acos3x9Bsin3xy_p''=-9A\cos3x-9B\sin3x

Substitute


9Acos3x9Bsin3x+15Asin3x15Bcos3x-9A\cos3x-9B\sin3x+15A\sin 3x-15B\cos3x

+6Acos3x+6Bsin3x=cos3x+6A\cos3x+6B\sin3x=\cos3x

3A15B=1-3A-15B=115A3B=0=>B=5A15A-3B=0=>B=5A

78A=1=>A=178,B=578-78A=1=>A=-\dfrac{1}{78},B=-\dfrac{5}{78}

Then


yp=178cos3x578sin3xy_p=-\dfrac{1}{78}\cos3x-\dfrac{5}{78}\sin3x

The general solution of the given nonhomogenepus equation is


y=c1e2x+c2e3x178cos3x578sin3xy=c_1e^{2x}+c_2e^{3x}-\dfrac{1}{78}\cos3x-\dfrac{5}{78}\sin3x


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