Help with ODE initial conditions

[ognik]

The ODE is $y'' + 4y' - 12y = 0$, I get $y = C_1e^{-6x} + C_2e^{2} $

The initial conditions are y(0) = 1, y(1)=2 - which gives me $C_1 = 1-C_2$ and $C_2 = \frac{2e^{6}-1}{e^{8}-1} $

This just looks more messy than book exercises normally are, and when I laboriously substitute back into the eqtn it doesn't resolve - so I think I have something wrong in my sltn, but just can't see it, could someone see what is wrong with my sltn please?

 

[MarkFL]

Well, the characteristic equation is:

\(\displaystyle r^2+4r-12=(r+6)(r-2)=0\)

Hence, the general solution is:

\(\displaystyle y(x)=c_1e^{-6x}+c_2e^{2x}\)

And we may compute:

\(\displaystyle y'(x)=-6c_1e^{-6x}+2c_2e^{2x}\)

Thus, the initial conditions result in the linear system:

\(\displaystyle c_1+c_2=1\)

\(\displaystyle -3c_1e^{-6}+c_2e^{2}=1\)

Solving this system, we obtain:

\(\displaystyle c_1=\frac{e^6\left(e^2-1\right)}{e^8+3},\,c_2=\frac{e^6+3}{e^8+3}\)

And thus the particular solution is:

\(\displaystyle y(x)=\frac{e^6\left(e^2-1\right)}{e^8+3}e^{-6x}+\frac{e^6+3}{e^8+3}e^{2x}\)

It appears you made a mistake solving for the parameters. :)

 

[ognik]

got it, thanks