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

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