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 September 21st, 2017, 06:35 PM #1 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Non-linear Non-separable first order differential equation Find the general solution of y' = (t + y - 1)^2 and write it in explicit form. This is clearly a non-linear, non-separable first order differential equation, but I'm really struggling; I don't know what method to use to solve this! What do you recommend? Last edited by skipjack; September 21st, 2017 at 09:21 PM.
 September 21st, 2017, 07:42 PM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,734 Thanks: 605 Math Focus: Yet to find out. This is Riccati's equation, a special case of the Bernoulli equation. Have you studied that? Thanks from topsquark, romsek and SenatorArmstrong Last edited by skipjack; September 21st, 2017 at 09:06 PM.
 September 21st, 2017, 09:19 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,866 Thanks: 1833 Let u = t + y - 1, then u' = 1 + y' = 1 + u², which is separable. Thanks from greg1313, topsquark and John Travolski
 September 22nd, 2017, 07:41 AM #4 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Yeah, that's the way to do it alright. Just not very intuitive, but that definitely works.
 September 22nd, 2017, 09:04 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra Spotting substitutions is almost as important in solving differential equations as it is in solving integrals. If that one isn't intuitive to you, I suggest you practice more. Thanks from greg1313 and topsquark
September 22nd, 2017, 11:07 PM   #6
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Quote:
 Originally Posted by John Travolski Yeah, that's the way to do it alright. Just not very intuitive, but that definitely works.
This is how I'd look at this problem.

$y^\prime = (y+t-1)^2$

as you noted it's not separable.

well what is separable? If you think like I do you'd come up with

$u^\prime = u^2$

and you can easily separate this into

$\dfrac{du}{u^2} = dt$

well given the form of the original equation what will $u$ have to be?

it's pretty clear that $u = y+t-1$

well ok let's try this and see what we get

$\dfrac{du}{dt} = \dfrac{dy}{dt} + 1$

$\dfrac{dy}{dt} = \dfrac{du}{dt} -1$

$u^\prime -1=u^2$

$u^\prime = u^2 + 1$

etc.

so sometimes you just have to try things out and see if they work.

September 23rd, 2017, 05:27 AM   #7
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Quote:
 Originally Posted by romsek sometimes you just have to try things out and see if they work.
This. In spades. Mathematics isn't about being to solve everything first time just by looking at it. It's about trying out the mathematical tools you have learned (or sometimes not). It's about applying techniques creatively to see what helps.

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