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[SOLVED] Help with ODE initial conditions

ognik

Active member
Feb 3, 2015
471
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Feb 3, 2015
471
got it, thanks