y''-2y'+y=0 ; With Solution y = e^x So far I got: let y=f(x)*v = e^xv y'=e^xv+e^xv' y''=e^xv+e^xv'+e^xv'+e^xv'' substituting that into y''-2y'+y=0 gets= e^xv'' and then v''=w' we get e^xw' not sure where to go from here though : (
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Originally Posted by Link88 y''-2y'+y=0 ; With Solution y = e^x So far I got: let y=f(x)*v = e^xv y'=e^xv+e^xv' y''=e^xv+e^xv'+e^xv'+e^xv'' substituting that into y''-2y'+y=0 gets= e^xv'' and then v''=w' we get e^xw' not sure where to go from here though : ( Since this is a second order linear constant coefficient ODE, find the characteristic equation , a repeated root. So your solution to the DE will have the form . You will need to use some intial conditions to find the constants.
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