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For what values of r does the function y = erx satisfy the differential equation y'' − 6y' + 2y = 0?...
4 months ago
Q:
For what values of r does the function y = erx satisfy the differential equation y'' − 6y' + 2y = 0? (enter your answers as a comma-separated list.)
Accepted Solution
A:
You mean y = e^(rx)? Yes, there end up being two values. They should both work. Nobody said the solution had to be unique!
Use this expression for y to calculated y', then take the derivative again to get y ' ' .
y' = r e^(rx) y' ' = r^2 e^(rx)
So you have: r^2 e^(rx) - 6r e^(rx) + 2e^(rx) = 0
Since e^(rx) can never be 0, we can divide both sides by e^(rx) to get
r^2 - 6r + 2 = 0
r1 =(6-√28)/2=3-√ 7 = 0.354
r2 =(6+√28)/2=3+√ 7 = 5.646
so r = 0.354, 5.646