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**Final Exam / Re: FE-P2**

« **on:**December 14, 2018, 09:28:44 AM »

I think Cui calculated the homogeneous solution wrong:

$(r-1)(r-1)(r-1) = r^3 -3r^2+3r+1$

Actually:

$r^3 -3r^2+4r-2= (r-1)(r^2-2r-2)$ = 0

$r=1$ or $r=1-i, 1+i$

So, Homogeneous solution is

$y= c_1e^t + c_2e^t\cos(t) + c_3e^t\sin(t)$

Non homogenous part for $10e^t$

We should assume $Y= Ate^t $

$Y’=Ate^t+Ae^t$

$Y’’=Ate^t+2Ae^t$

$Y’’’=Ate^t+3Ae^t$

$Y’’’-3Y’’+4Y’-2Y =(A-3A+4A-2A)te^t+(3A-6A+4A-2A)e^t =-Ae^t =10e^t$

$A =-10, Y=-10te^t$

$(r-1)(r-1)(r-1) = r^3 -3r^2+3r+1$

Actually:

$r^3 -3r^2+4r-2= (r-1)(r^2-2r-2)$ = 0

$r=1$ or $r=1-i, 1+i$

So, Homogeneous solution is

$y= c_1e^t + c_2e^t\cos(t) + c_3e^t\sin(t)$

Non homogenous part for $10e^t$

We should assume $Y= Ate^t $

$Y’=Ate^t+Ae^t$

$Y’’=Ate^t+2Ae^t$

$Y’’’=Ate^t+3Ae^t$

$Y’’’-3Y’’+4Y’-2Y =(A-3A+4A-2A)te^t+(3A-6A+4A-2A)e^t =-Ae^t =10e^t$

$A =-10, Y=-10te^t$