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# y''-4y'+3y=0 y(0)=0 and y'(0)=-1

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y''-4y'+3y=0
y(0)=0 and y'(0)=-1

Guest May 23, 2017
#1
+90052
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y''-4y'+3y=0             y(0)=0 and y'(0)=-1

We can solve this by letting the general solution be

y(t)  =  C1 e r1t  +  C2 e r2t

And  we can find r1  and r2 thusly

(r  - 3)  ( r - 1)  =  0         setting  both factors to 0 and solving for r  we have that r1   = 3  and r2  = 1

So......the general solution is :

y(t)  =  C1 e3t  +  C2 e1t      and

y' (t)  = 3Ci e 3t  +  C2 e 1t

Applying the initial conditions, we have that

0  =  C1  + C2     →    C2  =  - C1     (1)

-1  = 3C1  + C2       (2)

Subbing  (1)  into (2)  we have that

-1  =  3C1  -  C1

-1  =  2C1

-1/2  = C1    →    C2  = -C1  =  1/2

So......the actual solution is

y(t)  =  (-1/2) e 3t   +  (1/2) e1t

CPhill  May 23, 2017
edited by CPhill  May 23, 2017

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