# Correct Method for Second Order Separable Differential Equations

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 $$10dx^2$$ = $$10xdx$$?

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

$$\frac{d^2y}{dx^2} = 10$$ $$d^2y = 10 dx^2$$ 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 : $$y=5x^2+c_1x+c_2$$

• 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? – CanadianArcade Aug 31 '19 at 19:35
• 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 . – JJacquelin Aug 31 '19 at 20:57

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.

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 $$y(x)=y(x_0)+y'(x_0)(x-x_0)+5(x-x_0)^2.$$

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

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

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.