SOLUTION: Solve the initial-value problem d^(2)y/dt^(2)+4dy/dt+4y=0,y(1)=0,y′(1)=1. y(t)=

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Question 1184644: Solve the initial-value problem d^(2)y/dt^(2)+4dy/dt+4y=0,y(1)=0,y′(1)=1.
y(t)=
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!

The characteristic equation of the D.E. is lambda%5E2+%2B+4%2Alambda+%2B+4+=+0, which gives the double root lambda+=+-2.
===> The general solution to the D.E. is y+=+A%2Ae%5E%28-2t%29+%2B+Bt%2Ae%5E%28-2t%29.

==> %22y%27%281%29%22+=+%28-2A%2BB%29e%5E%28-2%29+-+2Be%5E%28-2%29+=1 ===> %28-2A-B%29e%5E%28-2%29+=+1 ===> 2A%2BB+=+-e%5E2.

Also, y%281%29+=+Ae%5E%28-2%29+%2B+Be%5E%28-2%29+=+0 ===> A + B = 0, or B = -A.

Combining this with the preceding equation, we get

2A+-+A+=+A+=+-e%5E2. ===> B+=e%5E2 ===> y+=+-e%5E2%2Ae%5E%28-2t%29+%2B+e%5E2%2At%2Ae%5E%28-2t%29.

after combining factors and simplifying, y%28t%29+=+%28t-1%29e%5E%28-2%28t-1%29%29