Question #302580

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


1
Expert's answer
2022-03-01T13:46:33-0500

y+4y=sin5x+x1y=y1+y21)y+4y=0λ2+4λ=0λ1=0,λ2=4y1=c1+c2e4x2)y2=acos5x+bsin5x+x(kx+p)y2=5asin5x+5bcos5x+2kx+py2=25acos5x25bsin5x+2k25acos5x25bsin5x+2k20asin5x++20bcos5x+8kx+4p=sin5x+x1cos5x:25a+20b=0sin5x:25b20a=1x:8k=1x0:2k+4p=1a=4205b=141k=18p=516y2=4205cos5x141sin5x+x(18x516)y=c1+c2e4x4205cos5x141sin5x+x(18x516)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})


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