Answer to Question #302579 in Differential Equations for haru

Question

Find the general solution of the following differential equations using method of undetermined coefficients: , (ii) 2y''−5y' = x^2 +5e^−4x,

Answer 1

Corresponding homogeneous differential equation

$2y''−5y' =0$

Characteristic (auxiliary) equation

$2r^2-5r=0$

$r_1=0,r_2=5/2$

The general solution of the homogeneous differential equation is

$y_h=c_1+c_2e^{5x/2}$

Find the particular solution of the non homogeneous differential equation

$y_p=Ax^3+Bx^2+Cx+De^{-4x}$

$y_p'=3Ax^2+2Bx+C-4De^{-4x}$

$y_p''=6Ax+2B+16De^{-4x}$

Substitute

$6Ax+2B+16De^{-4x}$

$-15Ax^2-10Bx-5C+20De^{-4x}$

$=x^2+5e^{-4x}$

$-15A=1$

$6A-10B=0$

$2B-5C=0$

$36 D=5$

$A=-1/15, B=-1/25, C=-2/125, D=5/36$

$y_p=-\dfrac{1}{15}x^3-\dfrac{1}{25}x^2-\dfrac{2}{125}x+\dfrac{5}{36}e^{-4x}$

The general solution of the non homogeneous differential equation is

$y=c_1+c_2e^{5x/2}-\dfrac{1}{15}x^3-\dfrac{1}{25}x^2-\dfrac{2}{125}x+\dfrac{5}{36}e^{-4x}$