❖ Reduction of Order: Basic Example in Differential Equations ❖

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Recommends John Wayland Bales · 2016-07 · Mathematics Stack Exchange · on “Legendre's equation of order n in differential equations”

Here is a good youtube explanation of the process of "reduction of order" you can use to find a second solution to a second order DE given a first solution. [this video]

Recommends xMathFanx · 2018-02 · Mathematics Stack Exchange · on “Is there any general method to solve a Second order ODE with variable coefficient of the form $a(x)y''(x)+b(x)y'(x)+c(x)y(x)=0$”

In order to successfully approach the problem submitted, the 'Reduction of Order' technique should prove quite useful. Here is a short video that will demonstrate how to solve such a problem with this method: [this video]

Recommends Moo · 2019-08 · Mathematics Stack Exchange · on “Reduction of Order Methods”

We are given $$\tag 1 t^2 y'' + ty' - y = 0, y_1 = t$$ This is a homogeneous linear equation and we can use many methods to find a general solution. For example, we can solve this problem as a homogeneous linear equation, exact equation or use a substitution method like Reduction of Order [LINK] . Method I: Solve as an Euler-Cauchy type equation. Assume $$y = t^m \implies y' = m t^{m-1} \implies y'' = m(m-1) t^{m-2}$$ Substituting into $(1)$ , we find that $m^2 - 1 = 0 \implies m = \pm~ 1$ , which leads to the general solution $$y = c_1 t + \dfrac{c_2}{t}$$ Method II: Solve using Reduction of Order We are told that a solution is $y_1 = t$ . We find the second solution using $$y_2 = v t \impl

Recommends user10478 · 2021-10 · Mathematics Stack Exchange · on “What determines how to treat single variable PDEs and thus their constants of integration?”

When solving a first-order ODE (perhaps there is also a way to extend this to a higher order ODE) for $y(x)$ , it is possible to shift perspective and consider $x$ to be a function of $y$ by manipulating the differential forms. For example, $\frac{dy}{dx} = h(y)$ can be solved as $\frac{dx}{dy} = \frac{1}{h(y)}$ , which may be easier. The moral of the story, very crudely stated, seems to be that derivatives are slopes, and we can inquire about the slope of any variable with respect to any other, regardless of which variables are "actually independent" at the end of the day. When solving a first-order PDE such as $u_t + 2xu_x = 0$ , one of the steps involves considering either $\frac{dt}{dx}$