Question #108726
y"−3y' + 2y = 3e−x−10cos3x, y(0) = 1, y'(0) = 2.
1
Expert's answer
2020-04-09T14:53:21-0400

The general solution of the second order nonhomogeneous linear equation can be expressed in the form


y=yc+Yy=y_c+Y

where YY is any specific function that satisfies the nonhomogeneous equation, and yc=C1y1+C2y2y_c=C_1y_1+C_2y_2 is a general solution of the corresponding homogeneous equation


y3y+2y=0y''-3y'+2y=0

The Characteristic Equation


r23r+2=0r^2-3r+2=0(r1)(r2)=0(r-1)(r-2)=0

r1=1,r2=2r_1=1, r_2=2

The general solution of the corresponding homogeneous equation is


yc=C1ex+C2e2xy_c=C_1e^{x}+C_2e^{2x}

Use Method of Undetermined Coefficients


Y=Aex+Bcos(3x)+Csin(3x)Y=Ae^{-x}+B\cos(3x)+C\sin(3x)


Y=Aex3Bsin(3x)+3Ccos(3x)Y'=-Ae^{-x}-3B\sin(3x)+3C\cos(3x)

Y=Aex9Bcos(3x)9Csin(3x)Y''=Ae^{-x}-9B\cos(3x)-9C\sin(3x)

Aex9Bcos(3x)9Csin(3x)+Ae^{-x}-9B\cos(3x)-9C\sin(3x)+

+3Aex+9Bsin(3x)9Ccos(3x)++3Ae^{-x}+9B\sin(3x)-9C\cos(3x)+

+2Aex+2Bcos(3x)+2Csin(3x)=3ex10cos(3x)+2Ae^{-x}+2B\cos(3x)+2C\sin(3x)=3e^{-x}-10\cos(3x)

6A=3=>A=126A=3=>A=\dfrac{1}{2}

7B9C=10-7B-9C=-10

9B7C=09B-7C=0

B=713,C=913B=\dfrac{7}{13} , C=\dfrac{9}{13}


Y=12ex+713cos(3x)+913sin(3x)Y=\dfrac{1}{2}e^{-x}+\dfrac{7}{13}\cos(3x)+\dfrac{9}{13}\sin(3x)


y=C1ex+C2e2x+12ex+713cos(3x)+913sin(3x)y=C_1e^{x}+C_2e^{2x}+\dfrac{1}{2}e^{-x}+\dfrac{7}{13}\cos(3x)+\dfrac{9}{13}\sin(3x)

y(0)=1:y(0)=1:

1=C1+C2+12+7131=C_1+C_2+\dfrac{1}{2}+\dfrac{7}{13}

y(0)=2:y'(0)=2:

y=C1ex+2C2e2x12ex2113sin(3x)+2713cos(3x)y'=C_1e^{x}+2C_2e^{2x}-\dfrac{1}{2}e^{-x}-\dfrac{21}{13}\sin(3x)+\dfrac{27}{13}\cos(3x)

2=C1+2C212+27132=C_1+2C_2-\dfrac{1}{2}+\dfrac{27}{13}

1=C21+20131=C_2-1+\dfrac{20}{13}

C2=613C_2=\dfrac{6}{13}


C1=12C_1=-\dfrac{1}{2}

yp=12ex+613e2x+12ex+713cos(3x)+913sin(3x)y_p=-\dfrac{1}{2}e^{x}+\dfrac{6}{13}e^{2x}+\dfrac{1}{2}e^{-x}+\dfrac{7}{13}\cos(3x)+\dfrac{9}{13}\sin(3x)




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