Finding Solutions to Second Order Differential Equations with Initial Conditions

[cemar.]

y'' - 8y' + 16 = 0 ; y(0) = 4 ; y'(0) = 9


this should be not too bad but I am stuck in the same place.
m^2 - 8m + 16
m1 = m2 = 4

y = C1e^(mx) + C2(e^(mx))
sub in y(0) = 4
4 = C1 + C2
C1 = 4-C2

y' = mC1e^(mx) + mC2e^(mx)
sub in y'(0) = 9
9 = mC1 + mC2
C1 = 4 - C2 (from above), m=4
9 = 4(4-C2) + 4C2
9 = 16 - 4C2 + 4C2
9 = 16
In this case would there just be no solution or am i missing something?!?
Thank you!

 

[nicksauce]

If I recall correctly, the general solution in the case of a repeated root for a second order constant coefficients ODE is
c_1e^{m_1x} + c_2xe^{m_1x}

 

[eys_physics]

Hey
The formula you are using for y is only valid if you have to different solutions m, i.e.
m_{1}\neq{m_{2}}

In your case y is given by

y=(C_{1}+C_{2}x)e^{mx}

 

[cemar.]

thanks!

 

[HallsofIvy]

Well done, guys.