Asked by Sandy
Solve the initial-value problem.
y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1
y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1
Answers
Answered by
Steve
y = e^x (c1 + c2*x)
y' = e^x (c1 + c2 + c2*x)
Now plug in the initial conditions to get
e^2 (c1 + 2c2) = 0
e^2 (c1 + c2 + 2c2) = 1
Now just solve for c1 and c2
y' = e^x (c1 + c2 + c2*x)
Now plug in the initial conditions to get
e^2 (c1 + 2c2) = 0
e^2 (c1 + c2 + 2c2) = 1
Now just solve for c1 and c2
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