# Solutions for $\frac{dy}{dx}=y$?

Al-right, this may be a very basic question but I'm confused about this. We all know that one differential equation can only have one solution. Consider:

$$\frac{dy}{dx}=y$$

The solution is:

$$y= e^x = c(1 + x +x^2/2 + ... ) = c( 1 + \int 1\times dx + \int \int 1 \times dx^2 + ...$$

where c is a constant. Wouldn't this work as a solution as well?

$$y = c(... -1/x^2 + 1/x + \log(x) + (x \log(x) - x) + ... = c \times (...+ \frac{d^2 (1/x)}{dx^2} + \frac{d (1/x)}{dx} + (1/x) + \int 1/x \times dx + ... )$$

Can someone tell me why the seond solution is wrong?

• How did you derive that second "solution"? Jul 15 '14 at 12:10
• I guessed it. One can verify it by: $$y = c \times (...+ \frac{d^2 (1/x)}{dx^2} + \frac{d (1/x)}{dx} + (1/x) + \int 1/x \times dx + ... )$$ Then $$dy/dx = c \times (...+ \frac{d^2 (1/x)}{dx^2} + \frac{d (1/x)}{dx} + (1/x) + \int 1/x \times dx + ... ) = y$$ Jul 15 '14 at 12:15
• @Hakim, Both solutions are sums of all derivatives and antiderivatives of some function $f$. In the case of $e^x$, $f=1$. In the other case OP took $f=1/x$. Jul 15 '14 at 12:16
• My guess is that the infinite sum of your solution diverges. Jul 15 '14 at 12:17
• So in essence I could possibly take $$y = ... f''(x) + f'(x) + f(x) + \int f(x) dx + ...$$ will always diverge unless $f(x) = c e^x$ ? Jul 15 '14 at 12:22

By `solution' you mean solution to the initial value problem (IVP) $$\frac{dy}{dx} = y,\qquad y(0) =c,\qquad x\in\ \mathbb{R}.$$
So if $y(x) = ce^x$ is a solution, then it is the only one. For the second 'solution', $\lim_{x\rightarrow 0}y$ is not even defined!