I find it's helpful to do a quick test of homogeneity: replace x with cx and y with cy, can you factor out all the c's and get rid of them? If you can, it's homogenous and you can do v = y/x.
so here are my questions what is separable and non-separable at @0:36 so I can understand. how do I know leaner or not leaner? i know this is basic question but I'm truly didn't understand the,
A linear equation only has first powers of y and its derivatives, and that these are not multiplied together. The y*y' term @1:23 makes the equation non-linear. The classification video goes into more detail: ruclips.net/video/xa64IcGDj1M/видео.html&ab_channel=TheLazyEngineer
So in theory, we could do other substitutions of this nature, right? Like you could conceivably have an equation where a substitution of v = y^2/cos(x) made things easier (somehow), and you could derive dv/dx from there and substitute in the same fashion. It's just that the "v = y/x" substitution lends itself to simplicity in ways that other substitutions do not. Am I following correctly?
... after fiddling with this, I think I can answer my own question. What makes homogenous equations so special in this regard is that dy/dx = v + x(dv/dx) -- that leads to the most separable situation possible, where the only appearances of "x" in the entire equation are already being multiplied together, and just need to shed the "dv" to achieve full separation. And you can get to that level of easy separability only if the product rule is operating on "vx", which leaves x-oriented terms of "1" and "x/dx". There might be special cases somewhere such that a substitution of "v=y^2/cos(x)" happens to simplify nicely, but those are special cases and not a strategy.
Replying to myself yet again, Bernoulli equations are conceptually similar: you can turn them linear just by doing v = y^(1-n), which just happens to make all the "v" terms cancel out except for one. So yes, once in a while there are known substitutions that take an equation from "unsolvable" to "manageable".
Thank you for all these excellent videos on Diff Eq. I really like how you recap previous strategies.
I find it's helpful to do a quick test of homogeneity: replace x with cx and y with cy, can you factor out all the c's and get rid of them? If you can, it's homogenous and you can do v = y/x.
so here are my questions what is separable and non-separable at @0:36 so I can understand. how do I know leaner or not leaner? i know this is basic question but I'm truly didn't understand the,
A linear equation only has first powers of y and its derivatives, and that these are not multiplied together. The y*y' term @1:23 makes the equation non-linear. The classification video goes into more detail: ruclips.net/video/xa64IcGDj1M/видео.html&ab_channel=TheLazyEngineer
Wow!! You make it so easy!!!
Great series! Can you explain why you only considered the positive square root in the last step?
Good catch! The negative sqrt should also work! It would have been more accurate if I wrote +/- sqrt(...) as you point out! Thank you.
So in theory, we could do other substitutions of this nature, right? Like you could conceivably have an equation where a substitution of v = y^2/cos(x) made things easier (somehow), and you could derive dv/dx from there and substitute in the same fashion. It's just that the "v = y/x" substitution lends itself to simplicity in ways that other substitutions do not. Am I following correctly?
... after fiddling with this, I think I can answer my own question. What makes homogenous equations so special in this regard is that dy/dx = v + x(dv/dx) -- that leads to the most separable situation possible, where the only appearances of "x" in the entire equation are already being multiplied together, and just need to shed the "dv" to achieve full separation. And you can get to that level of easy separability only if the product rule is operating on "vx", which leaves x-oriented terms of "1" and "x/dx". There might be special cases somewhere such that a substitution of "v=y^2/cos(x)" happens to simplify nicely, but those are special cases and not a strategy.
Replying to myself yet again, Bernoulli equations are conceptually similar: you can turn them linear just by doing v = y^(1-n), which just happens to make all the "v" terms cancel out except for one. So yes, once in a while there are known substitutions that take an equation from "unsolvable" to "manageable".
Do you mean homogeneous?
+Aaron Dixon no. This is something different. Not to be confused with homogeneous. See Boyce (text).