Find a general solution to y''+4y'+3.75y=109\cos5x To solve this, the first thing I did was find the general solutionto the homogeneous equivalent, and got c_1e^{-5x/2}+c_2e^{3x/2} Then i used the form K\cos(wx)+M\sin(wx) and got -2.72\cos(5x)+2.56\sin(5x) as a solution of the nonhomogeneous ODE

glamrockqueen7

glamrockqueen7

Answered question

2021-03-25

Find a general solution to y+4y+3.75y=109cos5x
To solve this, the first thing I did was find the general solutionto the homogeneous equivalent, and got
c1e5x2+c2e3x2
Then i used the form Kcos(wx)+Msin(wx) and got 2.72cos(5x)+2.56sin(5x) as a solution of the nonhomogeneous ODE

Answer & Explanation

Obiajulu

Obiajulu

Skilled2021-03-27Added 98 answers

Question:
Find a general solution to y+4y+3.75y=109cos5x
(D2+4D+3.72)Y=0
(D+2.5)(D+1.5)Y=0
C.F.=C1E2.5x+C2E1.5X
C.F.=C1E2.5X+C2E1.5X
To solve this, the first thing I did was find the general solutionto the homogeneous equivalent, and got c1e5x2+c2e3x2
Then I used the form Kcos(wx)+Msin(wx)
[1D2+4D+3.75]109cos(5X)=109[152+4D+3.75]cos(5X)
=109[14D+21.254D21.25(4D+21.25)]cos(5X)
=1094D+21.2516D221.252cos(5X)
=109[4D+21.25165221.252]cos(5X)
=10945sin(5X)+21.25cos(5X)400451.5625
=109[20sin(5X)21.25cos(5X)]851.5625
=2.56sin(5X)2.72cos(5X)
and got 2.72cos(5x)+2.56sin(5x) as a solution of the nonhomogeneous

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