solution for a first order ordinary differential equation with displacement I have to described the solutions for the following IVP
$$y'(t)=y(kt), y(0)=1$$
where $k$ is a positive constant. 
I tried to solved it but didn't get anywhere, can anybody give me a clue of how to solve this?
 A: Look for a solution as a power series
$$
y(t)=1+\sum_{n=1}^\infty a_n\,t^n.
$$
Then
$$
y'(t)=\sum_{n=1}^\infty n\,a_n\,t^{n-1}=\sum_{n=0}^\infty (n+1)\,a_{n+1}\,t^n
$$
and
$$
y(k\,t)=1+\sum_{n=1}^\infty a_n\,k^n\,t^n.
$$
Equating the coefficients of equal powers we get
$$
a_1=1,\quad a_{n+1}=\frac{k^n}{n+1}\,a_n,\quad n\ge1.
$$
This gives
$$
y(t)=1+\sum_{n=1}^\infty k^{n(n-1)/2}\frac{t^n}{n!}.
$$
Note: as observed in the comments, the series does not converge if $|k|>1$.
A: $$y(0)=1\\y'(t)=y(kt) \rightarrow y'(0)=y(0k)=y(0)=1\\y''(t)=ky'(kt) \rightarrow y''(0)=ky'(0)=k(1)=k\\$$ $$y'''(t)=k^2y''(kt) \rightarrow y'''(0)=k^2y''(0)=k^2(k)=k^3\\ $$ $$y''''(t)=k^3y'''(kt) \rightarrow y''''(0)=k^3y'''(0)=k^3(k^3)=k^6\\ y^{(5)}(t)=k^4y^{4}(kt) \rightarrow y^{(5)}(0)=k^4y^{4}(0) =k^4 *k^6 =k^{10}\\y^{(6)}(0)=k^{15}\\y^{(7)}(0)=k^{21}\\... $$ now :suppose $$y(t)=c_0 +c_1 t +c_2 t^2 +c_3t^3 +c_4 t^4 +...\\y(0)=c_0=1 \rightarrow c_0=1  \\y'(0)=c_1=1\\y''(0)=2!c_2=k \rightarrow c_2=\frac{k}{2!}\\y'''(0)=3!c_3=k^3 \rightarrow c_3=\frac{k^3}{3!}\\y^{(4)}(0)=4!c_4=k^6 \rightarrow c_4=\frac{k^6}{4!}\\...$$
