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Where is the flaw in my reasoning?

$$y''=y \iff y'=\frac{y^2}{2} \iff y=\frac{y^3}{6}$$

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  • $\begingroup$ Remember that you are dealing with $$\frac{d^2y(x)}{dx^2} = y(x) $$ $\endgroup$ Commented Feb 2, 2014 at 3:18
  • $\begingroup$ Even if your approach were right (see answers) you would be missing constants of integration. Claude's solution shows how you get two constants to be determined from boundary conditions. $\endgroup$ Commented Feb 2, 2014 at 6:33

2 Answers 2

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when you write $y''$ you mean differentiating with respect to some variable $x$.

However $\int y dy = y^2/2$

So when you go from $y''$ to $y'$ you are integrating with respect to $x$

So when you go from $y$ to $y^2/2$ you are integrating with respect to $y$

That is the flaw in your logic.

If you write out $$ \frac{d y^2}{d x^2} = y $$ then $$ \frac{d y}{d x} = \int y ~dx $$

For this reason alone, I am not a big fan of $y'$, $y''$ etc.

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    $\begingroup$ Leibniz notation is far from perfect. Why not just write $y'(x)$ instead? $\endgroup$
    – user71641
    Commented Feb 2, 2014 at 3:50
  • $\begingroup$ The problem is, it does not say what the independent variable is. For example, how will you interpret $y'(0)=2$? While I agree that Leibniz's notation also has ambiguities it at least shows what the independent varible is $\endgroup$
    – user44197
    Commented Feb 2, 2014 at 3:56
  • $\begingroup$ If you're working with a differential equation, then you'll be writing $y'(x)$ which makes it clear what the independent variable is. If you're working with expressions like $y'(3)$, then what the independent variable is should have been specified, but even if it hasn't it shouldn't matter since it's just a constant. $\endgroup$
    – user71641
    Commented Feb 2, 2014 at 4:13
  • $\begingroup$ no, only $\dot{y}$ is correct, both $\frac{dy}{dx}$ and $y'$ are wrong ;) $\endgroup$ Commented Feb 2, 2014 at 4:56
  • $\begingroup$ I still teach using a black board and I am not known to clean the black board all that well, so I don't like either $\dot{y}$ or $y'$ ;) $\endgroup$
    – user44197
    Commented Feb 2, 2014 at 5:26
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As said in previous answers, the problem is much clearer if you rewrite $$y''=y$$ as $$ \frac{d^2 y}{d x^2} = y$$ This really shows the dependencies between $y$ and $x$.

So, you face a second order differential equation the characteristic equation being $r^2=1$ which has two roots $r=1$ and $r=-1$; so its general solution write $$y=c_1 e^x+c_2 e^{-x}$$

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  • $\begingroup$ why not $\frac{d^2 y}{d x^2} = y$? $\endgroup$
    – dwarandae
    Commented Feb 2, 2014 at 6:43
  • $\begingroup$ @dwarandae. Thanks for reporting the typo. Fixed now. $\endgroup$ Commented Feb 2, 2014 at 7:15

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