y'' - 6y' + 5y = 0

try y = e^mx

we get the characteristic equation:

m^2 - 6m + 5 = 0

=> (m - 5)(m - 1) = 0

=> m = 5 or m = 1

so our general solution is of the form:

y = Ae^5x + Be^x.............general solution

now we have to find the particular solution with the conditions

y(0) = 0, y'(0) = 4

now if y = Ae^5x + Be^x

=> y' = 5Ae^5x + Be^x

since y(0) = 0 and y'(0) = 4, we have the simultaneous equations

Ae^0 + Be^0 = 0

5Ae^0 + Be^0 = 4

that is

A + B = 0 .............(1)

5A + B = 4 ............(2)

=> 4A = 4 .............(3) = (2) - (1)

=> A = 1

but A + B = 0

=> 1 + B = 0

=> B = -1

so the particular solution is:

y = e^5x - e^x...............particular solution

any questions?