Find the solution of the initial value problem y"-6y+5y=0;y(0),y"(0)?


2 Answers

Oddman Profile
Oddman answered
The solution is
  y(t) = a*e^t + b*e^(5t)
where "a" and "b" are constants that depend on the values of the initial conditions.   b = (y''(0) - y'(0))/20
  a = y'(0) - 5b
Forget Foxx Banda Profile
Your question is called a non-routine problem
A quick answer is true when both y"=0 and y=0
We can conclude the system by saying y"=y.
For any other answers where y">or<y the answer is not equal to 0

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