Existence of a twice differentiable function $f$ such that $f''-2f'+f=0$ and $f(0)=a,f'(0)=b$ for given values $a,b$

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?

• Do you know how to find the general solution to a linear, constant-coefficient differential equation? – carmichael561 Mar 29 '18 at 4:17
• Experiment with $t \mapsto e^{-t}$ and $t \mapsto t e^{-t}$. – copper.hat Mar 29 '18 at 4:36

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