7
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ How did you derive that second "solution"? $\endgroup$ – Hakim Jul 15 '14 at 12:10
  • $\begingroup$ 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 $$ dy/dx = c \times (...+ \frac{d^2 (1/x)}{dx^2} + \frac{d (1/x)}{dx} + (1/x) + \int 1/x \times dx + ... ) = y $$ $\endgroup$ – drewdles Jul 15 '14 at 12:15
  • $\begingroup$ @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 $f=1/x$. $\endgroup$ – Karolis Juodelė Jul 15 '14 at 12:16
  • 1
    $\begingroup$ My guess is that the infinite sum of your solution diverges. $\endgroup$ – Karolis Juodelė Jul 15 '14 at 12:17
  • $\begingroup$ 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 $ ? $\endgroup$ – drewdles Jul 15 '14 at 12:22
1
$\begingroup$

By `solution' you mean solution to the initial value problem (IVP) $$ \frac{dy}{dx} = y,\qquad y(0) =c,\qquad x\in\ \mathbb{R}.$$

It follows from the Picard–Lindelöf theorem there exists a solution to the above IVP and that solution is unique.
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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.