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Why can I just change $y''$ to $m^2$?

So for example: $$y''+y'-2y=0$$ $$m^2+ m -2=0$$ $$=(m+2)(m-1)$$ $m=-2,1$

$Ae^{-2x}+Be^x$

But where does that change come from?

$y''=m^2$ $y'=2m$ $y=2$ Is almost right? But we don't divide by $2$?

$m^2 + \frac m2 - 1$ makes more sense from above. Thanks guys!

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the change is, in your notation, $y''=m,y'=m,y=1$. The reason we do this in this case is we are assuming we have a solution of the form $y=e^{mx}$, then trying to find the appropriate values of m to make this work. So, $y'=me^{mx}$,$y''=m^2e^{mx}$. Plugging those in, we get $m^2e^{mx}+me^{mx}-2e^{mx}=0$. Now we factor, getting $e^{mx}(m^2+m-2)=0$. But since $e^{mx}$ is never equal to 0, that gives us $m^2+m-2=0$, which we then solve as a quadratic, getting your $m=1$ or $m=-2$. This tells us that any solution of the form $y=e^{-2x}$ or $y=e^x$ will work. What's more, if we throw a constant multiple against either of these, they will also work, because that constant multiple will just be carried through each of the derivatives, and then get factored out and cancelled. Plus, if we add 2 solutions of this form, we get another solution by the same reasonings (the derivative being a linear operator)

Hence, any solution of the form $y=Ae^{-2x}+Be^x$ will work.

Now, because this always works the same way, we stop writing the intermediate steps and just go straight into the characteristic equation, but this is the logic for where it came from

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You are looking for solutions of the form $y = A \mathrm{e}^{m x}$. Now observe that $y' = A m \mathrm{e}^{mx}$ and $y'' = A m^2 \mathrm{e}^{mx}$. Substitution yields

\begin{align*} A m^2 \mathrm{e}^{mx} + A m \mathrm{e}^{mx} - 2 A \mathrm{e}^{mx} = 0 \\ m^2 + m - 2 = 0 \end{align*}

In the second line I have divided by $A \mathrm{e}^{mx}$ which is allowed because for $A \neq 0$, $A \mathrm{e}^{mx}\neq 0, \forall x$.

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You usually assume a solution of the form $$y=e^{mx}$$ Then $$y''+y'-2y=0\Rightarrow m^2 e^{mx}+me^{mx}-2e^{mx}=0$$ Given that $e^{mx}>0$ then the above equation is equivalent to $$m^2+m-2=0\Rightarrow (m+2)(m-1)=0\Rightarrow m=1\hspace{0.5cm}or\hspace{0.5cm}m=-2$$ So the general solution would be $$y(x)=Ae^{x}+Be^{-2x}$$ The equation $m^2+m-2=0$ is called the auxiliary equation of the homogenous differential equation.

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