0
$\begingroup$

When considering a second order differential equation, say: $$\frac{d^2y}{dx^2} = 10$$ is it possible to separate and integrate such? Separating the variables results in: $$d^2y = 10 dx^2$$ and integrating results in: $$y dy = 10xdx$$and then integrating once again results in: $$\frac{y^2}{2} = 5x^2$$ Clearly solving for $y$ would yield something which is not the correct answer. I considered the scenario where the integral of $d^2y$ would be $dy$, but if that is the case why is the integral of $10dx^2$ = $10xdx$?

$\endgroup$
5
  • 1
    $\begingroup$ If $y''=10,$ then $y$ has the form $$5x^2+ax+b.$$ $\endgroup$ – Allawonder Aug 31 '19 at 7:13
  • $\begingroup$ @Allawonder I understand such, the issue I am having is why separation of variables will yield the incorrect answer. $\endgroup$ – CanadianArcade Aug 31 '19 at 7:24
  • $\begingroup$ It's because you tried integrating a second derivative. We can't treat second derivatives as fractions in any rigorous way. $\endgroup$ – Ninad Munshi Aug 31 '19 at 7:27
  • 2
    $\begingroup$ The $2$ in $d^2y$ is in a different place from the $2$ in $dx^2$. That's why they react differently to antidifferentiation. They're different! $\endgroup$ – Gerry Myerson Aug 31 '19 at 7:28
  • $\begingroup$ You might want to look at this question $\endgroup$ – Varun Vejalla Aug 31 '19 at 7:32
1
$\begingroup$

$$\frac{d^2y}{dx^2} = 10$$ $d^2y = 10 dx^2$ is non-sens because $\frac{d^2y}{dx^2}$ is not a fraction but is a conventional symbol meaning that the function $y(x)$ is differentiated two times successively.

A more comprehensive writing is : $$\frac{d}{dx}\left(\frac{dy}{dx}\right)=10$$ Then you can separate : $$d\left(\frac{dy}{dx}\right)=10dx$$ Which is integrated as : $$\frac{dy}{dx}=10x+c_1$$ $$dy=(10x+c_1)dx$$ And integrated again : $$y=5x^2+c_1x+c_2$$

$\endgroup$
2
  • $\begingroup$ Ah, this makes sense. I appreciate it a lot, thanks! Is there a reasoning why dy/dx can be treated as a separable fraction, however? $\endgroup$ – CanadianArcade Aug 31 '19 at 19:35
  • $\begingroup$ In a few words limited by the size of a comment, one cannot treat rigorously the basic of differential calculus. A key is the notion of infinitesimals conventionally noted $dx$ , $dy$, … which are smaller than any real number ( in absolue value of course). There are several methods to rigorously relate $dy$ and $dx$ to the notion of derivative of a function $y(x)$. en.wikipedia.org/wiki/Infinitesimal ; en.wikipedia.org/wiki/Differential_calculus . $\endgroup$ – JJacquelin Aug 31 '19 at 20:57
0
$\begingroup$

I think the "separable" concept does not apply to second order equations. I think you'd have a hard time getting a correct solution of $${d^2y\over dx^2}=y$$ by separation of variables.

$\endgroup$
0
$\begingroup$

The correct difference equation is $$ y(x+dx)-2y(x)+y(x-dx)=(d^2y)(x)=10(dx)^2 $$ This can be solved as any other difference equation. The left side has as characteristic roots $1$ as a double root, which is in resonance with the constant term on the right side, so you get as solution of the inhomogeneous difference equation $$ y(x+n\,dx)=A+Bn+Cn^2. $$ Inserting this for $n=-1,0,1$ gives \begin{align} A&=y(x),\\ 2B&=y(x+dx)-y(x-dx)=2y'(x)\,dx,\\ 2C&=y(x+dx)-2y(x)+y(x-dx)=10(dx)^2, \end{align} so that $$ y(x+n\,dx)=y(x)+y'(x)\,(n\,dx)+5\,(n\,dx)^2 $$ or $$ y(x)=y(x_0)+y'(x_0)(x-x_0)+5(x-x_0)^2. $$

$\endgroup$
0
$\begingroup$

Think of

$$ y'y'' = 10 y' $$

or

$$ \frac 12\frac{d}{dx}(y')^2 = 10\sqrt{(y')^2} $$

now separating variables

$$ \frac 12\frac{d\eta}{\sqrt{\eta}} = 10 dx $$

and after integration

$$ \sqrt \eta = 10 x + c_1 = y' $$

and finally

$$ y = 5 x^2+c_1 x+ c_2 $$

$\endgroup$
0
$\begingroup$

Let's proceed appropriately: We have $\mathrm d\mathrm d y=\mathrm{d^2}y=10\mathrm d x^2=(10\mathrm d x)\mathrm d x.$ Integrating gives $\mathrm d y=(10x+a)\mathrm d x.$ Integrating again gives $$y=5x^2+ax+b,$$ as wanted.

$\endgroup$

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.