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Find the value(s) of m such that y = e^(mx) is a solution to the differential equation: y'' -8y' +15y = 0
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Use the Characteristic equation to find the values of m, by finding the roots of the CE
m=3 or 5
thanks, what was your working?
The Characteristic equation is is\[m^2-8m+15\rightarrow (m-5)(m-3)\rightarrow m=3,m=5\] the solution is: \[y=C_{1}e^{3x}+C_2e^{5x}\]
thanks i just got stuck at this point \[m ^{2}−8m+15\] i had \[m ^{2}y−8my+15y\] nd completely forgot about the CE equation thanks!
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