Second order linear ODE not making sense...

Question

I am given:

$y''-3y'+2y=0$

$y(0)=1$

$y'(0)=2$

I know that $r_1=2$ and $r_2=1$

The solution therefore is:

$y(x)=C_1e^x+C_2e^{2x}$

Solving for initial values, I have:

$y(0)=C_1+C_2=1$

$y'(0)=C_1+2C_2=2$

Did I make a mistake somewhere?

Answer 1

Indeed the solution of $y''-3y'+2y=0$, before incorporating the initial data, is of form $$ y=c_1\mathrm{e}^x+c_2\mathrm{e}^{2x}. $$ The initial data enforce on $c_1$ and $c_2$ the following restrictions: $$ c_1+c_2=1,\,\,\,c_1+2c_2=2. $$ This is a system of two equations with two unknowns: $c_1$ and $c_2$.

You need to solve this system.

The solution is

$c_1=0,\,\,\,c_2=1$, and hence $y(x)=\mathrm{e}^{2x}$.