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When I was solving some differential equations, I asked myself the following: Is there a function has the following: $$y'=y+1$$ $$y''=y+1$$ $$y'''=y+1$$ $$......$$ $$......$$

If the initial value is $$y(0)=1$$

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    $\begingroup$ What about $y=2e^x-1$? $\endgroup$ – Galc127 Dec 14 '14 at 12:51
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    $\begingroup$ @Amzoti how does this only work for the first one? For any $k$, $y^{(k)} = 2e^x$ and $y(0) = 1$ so I think this one work. $\endgroup$ – servabat Dec 14 '14 at 13:01
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Well you have that $$y'=y''=y^{(3)}\cdots$$ and the only function that is a derivative of itself is $ae^{x}$ for some $a$ so $y'=ae^x$ and $$y=y'-1=ae^x-1$$ Since $y(0)=1$ we have that $$y(0)=ae^0-1=a-1=1$$ so $a=2$ hence our function is $\boxed{y(x)=2e^x-1}$

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    $\begingroup$ +1, although the edge case mentioned doesn't need special consideration, since the function $y'=0$ is of the form $y'=ae^x$ with $a=0$. $\endgroup$ – Peter Olson Dec 14 '14 at 15:12
  • $\begingroup$ @PeterOlson Yes you're right,corrected,Thanks $\endgroup$ – kingW3 Dec 14 '14 at 21:57
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All your differential equations after the first one are implied by the first equation. If $y'=y+1$ then differentiation of it gives $y''=y'$ and further differentiation gives the successive equations. So it is really just about solving the first equation with the given initial value, and this you can do.

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The solution to $$ y' = y + 1,\ y(0)=1$$ is $y = 2e^x -1$ which can be found from your method of choice for first-order non-homogeneous ODEs. Now let's see what the derivatives look like

$$y' = 2e^x$$

$$y'' = 2e^x$$

$$y''' = 2e^x$$

We can clearly see $y^{(k+1)} = y^{(k)}\ \forall k\in\mathbb{Z},k\geq1 $. Therefor all $n$th-derivatives of $y$ are equal by induction, and as $y'$ satisfies the conditions so do all $y^{(k)}$.

Edit:

If you're interested, here's how you could solve this using annihilators, my method of choice for simple problems like this (I prefer Laplace for harder ones). First re-arrange the equation to say:

$$y'-y=1$$

We find the general solution, $y_c$ by setting the RHS to $0$:

$$y_c'-y_c = 0$$

For which we assume $y_c = Ce^{mx} \Rightarrow mCe^{mx}-Ce^{mx} = 0 \Rightarrow m = 1$ so therefor $y_c = Ce^x$. We need to find $y_p$ now.

As $D$ annihilates $1$, assume $y_p = a$. Then $y' = 0$ and thus $0 - a = 1$ so $a=-1$. All together we have $y = Ce^x-1$. Further

$$y(0) = Ce^0 - 1 = 1 \Rightarrow C = 2 \therefore y= 2e^x -1$$

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