Initial value problem of a second ODE.

y'' - y' -12y = 0.
I got the general solution.
y = c1*e^(4x) + c2*e^(-3x)

initial value are:
y(0) = 1, y'(0) = 4

How do you use these to get the constants? please help

Question

Initial value problem of a second ODE.

y'' - y' -12y = 0.
I got the general solution.
 y = c1*e^(4x) +  c2*e^(-3x)

initial value are:
y(0) = 1, y'(0) = 4

How do you use these to get the constants? please help

anonymous · Answer

so you know y(0)=1, so plugging in x = 0 into your general solutions gives:$$y(0)=c_1e^{4(0)}+c_2e^{-3(0)}\iff 1=c_1+c_2$$Take the first derivative of your general solution, we obtain:$$y'=4c_1e^{4x}-3c_2e^{-3x}$$Now, plug in x = 0 and we get:$$y'(0)=4c_1e^{4(0)}-3c_2e^{-3(0)}\iff 4=4c_1-3c_2$$So now we have this system of equations:$$c_1+c_2=1$$$$4c_1-3c_2=4$$Multiplying the first equation by 3 and adding them gives us:$$7c_1=7\iff c_1=1$$Thus we also get:$$c_2=0$$So your particular solution according to your initial conditions is:$$y_p=e^{4x}$$

anonymous · Answer

Thank you!