Answer to Question #273163 in Differential Equations for Adnan

Question

y''-8y'+20y=100x²-26xe^x solve the given differential equation by undetermined cofficient

Answer 1

The corresponding homogeneous differential equation

$$y''-8y'+20y=0$$

The auxiliary equation

$$r^2-8r+20=0$$

$$(r-4)^2=-4$$

$$r=4\pm2i$$

The general solution of the corresponding homogeneous differential equation is

$$y_h=c_1e^{4x}\cos(2x)+c_1e^{4x}\sin(2x)$$

Find the particular solution of the nonhomogeneous differential equation

$$y_p=Ax^2+Bx+C+(Dx+E)e^x$$

$$y_p'=2Ax+B+(Dx+E+D)e^x$$

$$y_p''=2A+(Dx+E+2D)e^x$$

Substitute

$$2A+(Dx+E+2D)e^x$$

$$-8(2Ax+B+(Dx+E+D)e^x)$$

$$+20(Ax^2+Bx+C+(Dx+E)e^x)$$

$$=100x^2-26xe^x$$

$$x^2:20A=100=>A=5$$

$$x^1:-16A+20B=0=>B=4$$

$$x^0:2A-8B+20C=0=>C=\dfrac{11}{10}$$

$$xe^x:D-8D+20D=-26=>D=-2$$

$$e^x:E+2D-8E-8D+20E=0=>E=-\dfrac{12}{13}$$

Then

$$y_p=5x^2+4x+\dfrac{11}{10}-2xe^x-\dfrac{12}{13}e^x$$

The general solution of the given nonhomogeneous differential equation is

$$y=c_1e^{4x}\cos(2x)+c_1e^{4x}\sin(2x)$$

$$+5x^2+4x+\dfrac{11}{10}-2xe^x-\dfrac{12}{13}e^x$$