I am looking at the following ODE (obtained in order to use the similarity reduction method when solving the Heat Equation):$$ y'' + \frac{1}{2}xy'=0 $$IC/BC: $y(0) = 1, y(\infty) = 0$
The text says: "Separating the variables, we find that:$$y'(x) = Ce^{-x^2/4}$$ for some constant $C$."
How do I go about getting the result presented above?
In particular I am confused as this is not the separation of variables (s.o.v) I have seen before e.g.:
- in the 1st order ODE where you integrate out both sides
- in the Heat Equation solution where the ratio of both $X$ and $T$ functions is the same on both sides of equation hence they must equal to the constant $\lambda$.
2 Answers
$\begingroup$$$y'' + \frac{1}{2}xy'=0$$The DE is linear... The integrating factor is:$$\mu=e^{\frac {x^2}{4}}$$$$y''e^{\frac {x^2}{4}} + \frac{1}{2}xe^{\frac {x^2}{4}}y'=0$$$$\left(y'e^{\frac {x^2}{4}} \right)'=0$$Integrate.
$\endgroup$ 2 $\begingroup$Supposing that $y'\not\equiv 0$, make the change of variables $u = y'$ so that we get\begin{align*} u' + \frac{xu}{2} = 0 & \Longleftrightarrow u' = -\frac{xu}{2}\\\\ & \Longleftrightarrow \frac{u'}{u} = -\frac{x}{2}\\\\ & \Longleftrightarrow \ln|u| = -\frac{x^{2}}{4} + c\\\\ & \Longleftrightarrow u = \pm\exp(c)\exp\left(-\frac{x^{2}}{4}\right)\\\\ & \Longleftrightarrow u = C\exp\left(-\frac{x^{2}}{4}\right)\\\\ & \Longleftrightarrow y' = C\exp\left(-\frac{x^{2}}{4}\right) \end{align*}
where $C\in\textbf{R}\backslash\{0\}$. Since $y'\equiv 0$ is also a solution, we can consider that $C\in\textbf{R}$.
Hopefully this helps !
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