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Does there exist a twice differentiable $f$ such that $f''-2f'+f=0 $ and $f(0)=a, f'(0)=b$ for given values $a,b$?

My attempt: By the linear relation given we get that$ f$ has derivatives of all orders, so the question boils down to whether there is a function whose derivatives of all orders are prescribed at a given point,here,$0$. If I consider the Taylor series, will that function be convergent?

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  • $\begingroup$ Do you know how to find the general solution to a linear, constant-coefficient differential equation? $\endgroup$ – carmichael561 Mar 29 '18 at 4:17
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    $\begingroup$ Experiment with $t \mapsto e^{-t}$ and $t \mapsto t e^{-t}$. $\endgroup$ – copper.hat Mar 29 '18 at 4:36
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Try substituting $f(x)=Ae^{kx}$ as a solution and we get:

$$k^2 (Ae^{kx}) -2k(Ae^{kx})+Ae^{kx}=0$$ $$\Rightarrow (k-1)^2=0$$ $$k=1$$ Thus, $Ae^{x}$ is a solution

Now try substituting $f(x)=Axe^{kx}$, to get $$(Axk^2e^{kx}+2kAe^{kx})-2(Akxe^{kx}+Ae^{kx})+Axe^{kx}=0$$ $$x(k^2-2k+1)+(2k-2)=0$$ $$\Rightarrow k=1$$ Thus, $f(x)=Axe^{x}$ is another solution. As a result, we have the general solution of the form $$f(x)=Ae^x +Bxe^x$$

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  • $\begingroup$ $x^2-2x+1 = (x-1)^2$. Nothing complex. $\endgroup$ – copper.hat Mar 29 '18 at 4:39

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