(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

initial conditions y(0)=0, y'(0)=1,y''(0)=1

2. Relevant equations

Assume [itex]y=e^{rt}[/itex]

3. The attempt at a solution

By the substitution I'm left with

[itex]r^3-3r+2=0[/itex]

which gives me the roots of -2 and 1.

my question is a lot of times with this type of question I can get three roots and get a genral solution of

[itex]y=c_1e^{rx}+c_2e^{rx}+c_3e^{rx}[/itex]

this time my solution is

[itex]y=c_1e^{-2x}+c_2e^{t}[/itex]

therefore

[itex]y'=c_1-2e^{-2x}+c_2e^{t}[/itex]

and

[itex]y''=c_14e^{-2x}+c_2e^{t}[/itex]

because I've only got the two constants I solved for [itex]c_1=\frac{-1}{3}[/itex] and [itex]c_2=\frac{1}{3}[/itex]

The problem is when I substitue back into the solutions for y,y',y''

I'm getting

initial conditions y(0)=0, y'(0)=1 which is good but I'm getting y''(0)=-1.

can some please let me know where I'm going wrong?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: General solution of third order DE

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