How to solve this non-linear differential equation

Подписаться 307 тыс.

Просмотров 10 тыс.

50% 1

🌟Support the channel🌟
Patreon: / michaelpennmath
Channel Membership: / @michaelpennmath
Merch: teespring.com/...
My amazon shop: www.amazon.com...
🟢 Discord: / discord
🌟my other channels🌟
mathmajor: / @mathmajor
pennpav podcast: / @thepennpavpodcast7878
🌟My Links🌟
Personal Website: www.michael-pen...
Instagram: / melp2718
Twitter: / michaelpennmath
Randolph College Math: www.randolphcol...
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...
🌟How I make Thumbnails🌟
Canva: partner.canva....
Color Pallet: coolors.co/?re...
🌟Suggest a problem🌟
forms.gle/ea7P...

Опубликовано:

17 окт 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 29

@aadfg0 2 месяца назад

Much faster: because RHS is complicated, let u = xy' - y. Voila, we get x^2(u' + 2) = u^2. Set v = u/x, now v'x + v + 2 = v^2 -> v' = (v-2)(v+1)/x, finish with separation of variables.

@franolich3 2 месяца назад

An alternative method which is perhaps slightly messier but more direct I think... x^3.y" + 2x^2 = (xy' - y)^2 => x^2.(xy" + 2) = x^2.(y' - y/x)^2 => xy" + 2 = (y' - y/x)^2 [eq.1] Let u = y' - y/x [eq.2] => u' = y" - y'/x + y/x^2 = y" - u/x => y" = u' + u/x [eq.3] [eq.2] and [eq.3] in [eq.1]: => x.(u' + u/x) + 2 = u^2 => xu' = u^2 - u - 2 => xu' = (u+1)(u-2) => 1/x = u'/(u+1)(u-2) => 3/x = u'[1/(u-2) - 1/(u+1)] => 3.ln|x| + ln|A| = ln|u-2| - ln|u+1| => ln|Ax^3| = ln|(u-2)/(u+1)| => Ax^3 = (u-2)/(u+1) => u - 2 = Ax^3.u + Ax^3 => u = - (Ax^3 + 2)/(Ax^3 - 1) => y' - y/x = - (Ax^3 + 2)/(Ax^3 - 1) Use integrating factor of 1/x: => y'/x - y/x^2 = - (Ax^3 + 2)/x.(Ax^3 - 1) => (y/x)' = 2/x - 3Ax^2/(Ax^3 - 1) => y/x = 2ln|x| - ln|Ax^3 - 1| + ln|B| => y = x.ln|Bx^2/(Ax^3 - 1)|

@redrosin99 2 месяца назад

I like how you chose the substitution function without limitation and then created a limitation that helped you solve the original equation. The substitution was both easy to solve on its own restriction and reduced unpleasant parts from both sides of the equation. Is it a practical procedure in the general sense?

@minamagdy4126 2 месяца назад

It's indeed great to see how such a substitution can be derived. In a way, (almost) all substitutions can be done this way, it's just that we tend to have an intuition for where some simple substitutions work (exhibit A: the first substitution here?). Also, while this generalizes, sometimes the functional equation or resulting DE tends to be gnarly in general.

@ethanbartiromo2888 2 месяца назад

That was a beautiful solution!

@zeychin 2 месяца назад

Always look forward to a new video from you and I've caught it fresh as can be. Thank you for your dedication to teaching!

@nickmulgan7771 2 месяца назад

maybe clearer motivation: xy'-y looks like the numerator of (y/x)'

@edmundwoolliams1240 2 месяца назад

12:42 No! That's not a good place to stop! What about the uniqueness of the solution, given the choice of f(u) to equate two terms and simplify?

@ryanmuller9497 2 месяца назад

This took me a bit to process as well. The key is that the choice of f(u) is basically arbitrary because u is also a function of x. Whatever f(u) we choose, there is some u(x) that can be composed with it to form the z(x) (and thus y(x)) that we're seeking. To use a simpler example, imagine that we knew z(x)=x. Now, if we chose f(u)=ln|u|, we would then know that u(x)=e^(x). If, instead, we chose f(u)=e^(u), we would then have u(x)=ln|x|. In our actual problem, we have a more complicated z(x), but the same principle applies: f(u) is arbitrary because u(x) will just become whatever function composes with the chosen f(u) to form z(x). So, that leaves us free to choose the most convenient f(u) to simplify the equation, and then solve what remains to find the necessary corresponding u(x). In theory, if you wanted to, you should be able to choose an alternative f(u), use the z(x) found in the solution to calculate u(x) for the new f(u), and show that the differential equation is still satisfied for the different f(u) and u(x) as long as they compose to the same z(x). Might make for a decent homework exercise to satisfy yourself that f(u) truly is arbitrary.

@edmundwoolliams1240 2 месяца назад

Thank you! Yeah, it's like an extra degree of freedom was added to the problem, and he just chose the most convenient value of that to make it as easy as possible to solve. Very clever. A classic case of making the problem more complicated to make it easier 😊

@holyshit922 2 месяца назад

4:21 Substitution u=z' gives us Riccati equation This Riccati equation has particular solution in the form u = a/x

@Keithfert490 2 месяца назад

Once you got to x^2 z'' + 2 x z' + 2 = x^2 (z')^2 I substituted w = x z' to achieve a first order separable DE for w.

@holyshit922 2 месяца назад

I solved this equation as Riccati but your solution is faster

@burk314 2 месяца назад

I didn't jump straight to your w = xz', but I did lower the order with a w=z' type substitution.

@Keithfert490 2 месяца назад

@@burk314 yeah, to be honest, I also did it in two substitutions lol. But easier to present in youtube comments this way

2 месяца назад

from Morocco...thank you for this complete clear solution

@yaseengarehmohammadlou9349 2 месяца назад

If you do a factorization of x square from the LHS, then you you can do a substitution xy'-y=u which if you take a derivative, it will appear in the LHS

@MortezaSabzian-db1sl 2 месяца назад

Well, if you are looking for an interesting solution, follow my comment First, I write the main equation x³y''+2x²=(xy'-y)² x²(xy''+2)=(xy'-y)² Let me u=xy''+2 and v=xy'-y So I turned a differential equation into a differential equation machine (system) Now I rewrite all my equations twice 1)x²u=v² 2)u=xy''+2 3)v=xy'-y Look at the third equation, I want to convert the right side of it into a sentence I think the way to do it is clear v/x²=y'/x-y/x²=(y/x)' Now let again y/x=z and then inserting the new parameter into the equations 1)x²u=v² 2)u=x(zx)''+2 3)v/x²=z' I use the third equation to remove his parameter in the first equation 1)u=(xz')² 2)u=x(zx)''+2 The two equations are equal (xz')²=x(zx)''+2 I calculate one of the two derivatives on the right side of the equation (xz')²=x(xz'+z)'+2=x(xz')'+xz'+2 Let and set xz'=t t²=xt'+t+2 Well, we have a separable equation that is easy to solve dt/(t²-t-2)=dx/x d(ln((t-2)/(t+1))=d(ln(Ax³)) (t-2)/(t+1)=Ax³ The process of isolating the t parameter t=3Ax³/(1-Ax³)+2 Placement of initial parameters z'=3Ax²(1-Ax³)+2/x dz=d(-ln(1-Ax³)+2ln(x)) z=ln(Bx²/(1-Ax³))=ln(x²/(C+Dx³) and finally y=xln|x²/(Dx³+C)|

@jimschneider799 2 месяца назад

Solved this during a rather boring meeting. Once it's done, I'll watch the video. I made the substitution x*t = x*y' - y, which gave x*y'' = x*t' + t. Further, I determined that if x*t = x*y' - y, then y = x*INT(t/x, x). Making the substitutions for x*y'' and x*y' - y turned the equation into x^2*(x*t' + t) + 2*x^2 = x^2*t^2. If x is equal to zero, the equation is vacuously true, so I assumed that x != 0. Dividing through by x^2 and rearranging gave me x*t' + t - t^2 = - 2. The form of the equation suggested to me that t is a polynomial, and solving for its coefficients very quickly resulted in the conclusion that either t = -1 or t = 2 are the only polynomial solutions. Solving the integral for y gave either y = x*(C - ln(x)) or y = x*(C + 2*ln(x)); substituting the more general y = x*(C + A*ln(x)) resulted in the conclusion that A can only be -1 or 2. I have not yet ruled out t = e^f(x), but I doubt it would work. If I find a solution of exponential form, I'll leave another comment.

@rainerzufall42 2 месяца назад

For the understanding of the last line, it would be better to write f(u) = - ln|u| = ln(1/|u|) Are A and B really free in IR ?

@burk314 2 месяца назад

Try substituting back into the original equation to see if there's any As or Bs that wouldn't work. It will be some quotient rules and algebra, but doable.

@Alan-zf2tt 2 месяца назад

Faced with such a differential equation I'd originally consider the brilliance of Messrs Cauchy and Euler then take up fishing instead of mathing 🙂 EDIT: I have not tried this (far too lazy today - a lazy Sunday) but it looks interesting. On other hand may be a blind alley? set x = θ and y(x) to y(θ) = e^(iθ) then y' = ie^(iθ) and y'' = iie^(iθ) = that is y'' = -y Plus! when it looks like many stages of differentiation it suggest a polynomial of powers greater than three or infinite degree OR angles such as sine cosine OR sinh and cosh (OR even more exotic ones Michael explained in recent months eg squine and cosquine) but I am just speculating on these after thunking about them overnight

@voxclamantisinextermo 2 месяца назад

Nice eq. I suggest next solve y''x^3+2x^2=(y'x-y)^2 y''x^3+2x^2 = x^4 ((y/x/)')^2 y/x = t, y = tx, y' = t'x+t, y'' = t''x+2t'; x^2t''+2t'x+2 = x^2(t')^2 x^2t''+3 = (xt'-1)^2 xt' = v, t' = v/x, t'' = v'/x - v/x^2 xv' - v + 3 = (v-1)^2 xv' = v^2v-v-2 -- Separable eq integral(dv/(v^2-v-2)) = integral(dx/x) ln((v-2)/(v+1))/3 = ln(x) + C =ln(xC_1) v = 3/(1-x^3C1) -1 t' = 3/x(1-x^3C1) -1/x t = 2ln(x) - ln(1-x^3C_1)+C_2 y = x(2ln(x) - ln(1-x^3C_1)+C_2 C_2 = ln(C_3) y = xln(C_3x^2/(1-x^3C_1))

@xizar0rg 2 месяца назад

Why can you suppose that u(x) is a polynomial in x?

@thelocalsage 2 месяца назад

Notice how the diff eq for u(x) has a coefficient of degree “n” for every term that is the n-th derivative of u(x). I.e. the 2nd derivative term has a coefficient that is a degree-2 term of x, the 1st derivative has a coefficient of degree-1, 0th derivative a coefficient of degree-0. Now notice how it’s all set to “0” on the other side. For this to be true, each term has to cancel with each other, so whatever u(x) is probably has a derivative that when multiplied by x takes the same form as u(x), and likewise for the second derivative times x^2. This general property is satisfied by polynomials because of the power rule, so assuming u(x) is a polynomial of x allows a particular solution that can then be abstracted to a more general solution.

@moonwatcher2001 2 месяца назад

❤and that's a good place to stop

@Etothe2iPi 2 месяца назад

I don't see why f'(u) is not -u'/u, since u is a function of x and ' clearly means differentiation with respect to x.

@declandougan7243 2 месяца назад

Why *would* it be -u’/u?

@burk314 2 месяца назад

When he is writing f' and f'' he means the derivatives with respect to u, not x. It's one of the drawbacks of the prime notation for derivatives, but it comes up when doing chain rule stuff. He set z = f(u) and so when differentiating with respect to x, he ends up with dz/dx = df/du * du/dx. He wrote dz/dx = z' and du/dx = u' so those primes mean with respect to x, but df/du = f'(u) so that prime is respect to u.