+0  
 
0
559
1
avatar

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

Guest May 23, 2017
 #1
avatar+90052 
+1

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

 

 

cool cool cool

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

9 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.