Answer to Question #302580 in Differential Equations for haru

Question

Find the general solution of the following differential equations using method of undetermined coefficients: (iii) y''+ 4y' = sin5x+x−1,

Answer 1

$y''+4y'=\sin 5x+x-1\\ y=y_1+y_2\\ 1) y''+4y'=0\\ \lambda^2+4\lambda=0\\ \lambda_1=0, \lambda_2=-4\\ y_1=c_1+c_2 e^{-4x}\\ 2) y_2=a \cos 5x+ b \sin5x +x(kx+p)\\ y'_2 =-5a \sin5x +5b \cos5x +2kx+p\\ y_2''=-25a \cos5x-25 b\sin5x+2k\\ -25a \cos5x-25 b\sin5x+2k -20a \sin5x +\\ +20b \cos5x +8kx+4p=\sin 5x+x-1\\ \cos5x: -25a+20b=0\\ \sin5x: -25b-20a=1\\ x: 8k=1\\ x^0: 2k+4p=-1\\ a=-\frac{4}{205}\\ b=-\frac{1}{41}\\ k=\frac{1}{8}\\ p=-\frac{5}{16}\\ y_2=-\frac{4}{205} \cos 5x-\frac{1}{41} \sin5x +x(\frac{1}{8}x-\frac{5}{16})\\ y=c_1+c_2 e^{-4x}-\\ -\frac{4}{205} \cos 5x-\frac{1}{41} \sin5x +x(\frac{1}{8}x-\frac{5}{16})$