Find values of c such that y=e^(cx) is a solution to the differential equation:

y''-3y'+2y=0

Question

Find values of c such that y=e^(cx) is a solution to the differential equation:

y''-3y'+2y=0

dumbcow · Answer

set up characteristic equation ... which is quadratic where y'' is squared term , y' is "x" term and y is contant

--> c^2 -3c +2 = 0

now solve for c

anonymous · Answer

Thanks, dumbcow. I had a feeling that the answer was found by using solving for a quadratic equation. So, how does find the zero's of the y function have anything to do with the y=e^x function?

anonymous · Answer

By y function I mean y'' -3y' +2y=0. Does it just so happen that the natural exponential function and this y'' -3y' +2y are related?

dumbcow · Answer

yeah it just so happens that the zeros are also the coefficients in y=e^cx here is a link that will help explain http://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx

anonymous · Answer

Thanks dumbcow, I appreciate all the help!

dumbcow · Answer

yw

anonymous · Answer

show the steps involved in working out the problem