Answer to Question #302578 in Differential Equations for haru

Question #302578

Find the general solution of the following differential equations using method of undetermined coefficients: (iii) y''− 2y'+y = xsinx.


1
Expert's answer
2022-02-28T16:11:22-0500

Corresponding homogeneous differential equation


"y''\u22122y'+y =0"

Characteristic (auxiliary) equation


"r^2-2r+1=0"

"r_1=r_2=1"

The general solution of the homogeneous differential equation is


"y_h=c_1e^x+c_2xe^x"

Find the particular solution of the non homogeneous differential equation


"y_p=Ax\\sin x+Bx\\cos x+C\\sin x+D\\cos x"

"y_p'=A\\sin x+Ax\\cos x+B\\cos x-Bx\\sin x"

"+C\\cos x-D\\sin x"

"y_p''=2A\\cos x-Ax\\sin x-2B\\sin x"

"-Bx\\cos x-C\\sin x-D\\cos x"

Substitute


"2A\\cos x-Ax\\sin x-2B\\sin x"

"-Bx\\cos x-C\\sin x-D\\cos x"

"-2A\\sin x-2Ax\\cos x-2B\\cos x+2Bx\\sin x"

"-2C\\cos x+2D\\sin x+Ax\\sin x+Bx\\cos x"

"+C\\sin x+D\\cos x=x\\sin x"

"2B=1"

"-2A=0"

"-2B-C-2A+2D+C=0"

"2A-D-2B-2C+D=0"


"A=0, B=1\/2, C=-1\/2, D=1\/2"

"y_p=\\dfrac{x}{2}\\cos x-\\dfrac{1}{2}\\sin x+\\dfrac{1}{2}\\cos x"

The general solution of the non homogeneous differential equation is


"y=c_1e^x+c_2xe^x+\\dfrac{x}{2}\\cos x-\\dfrac{1}{2}\\sin x+\\dfrac{1}{2}\\cos x"




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