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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?

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    $\begingroup$ What makes you think you have made a mistake? $\endgroup$ – Gerry Myerson Feb 28 '14 at 22:20
  • $\begingroup$ Probably because I expected an explicit solution. $\endgroup$ – westhe32nd Feb 28 '14 at 22:21
  • $\begingroup$ Well, you haven't finished --- you stopped when you got 2 equations in the 2 unknowns. Solve that system, and put the solution back into your formula for $y(x)$. $\endgroup$ – Gerry Myerson Feb 28 '14 at 22:22
  • $\begingroup$ The only thing not making sense right now is why you think something doesn't make sense. It's quite an easy set of equations to solve. $\endgroup$ – Mike Feb 28 '14 at 22:25
  • $\begingroup$ Just substitute in original DE and conditions $\endgroup$ – Semsem Feb 28 '14 at 22:25
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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 $y(x)=\mathrm{e}^{2x}$.

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  • $\begingroup$ I literally just solved this a few seconds ago before you posted. Thank you! $\endgroup$ – westhe32nd Feb 28 '14 at 22:43

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