How to solve exact and non-exact ODE

In this video , I explained how to make a non-exact ODE exact and the steps to complete the solution
• Exact Equations [ODE]

[@stevehughes7789]

I have not studied calculus or dif eq in thirty years, but I still watch all of your videos. I love your channel. Great work!!!

[@HaliPuppeh]

This is the first time I've seen someone CLEARLY explain how the heck to get the integrating factor. The DiffEq book that I have says that you can use one, but it doesn't tell you AT ALL where they pull it out of...

[@joelmacinnes2391]

Yeah I saw blackpenredpen explain it well, when we were taught in high school they just said we had to know what it was not how to find it, and at university level they just brushed over it in a slide - yes explaining it but not giving it even 10 seconds to set in

[@joelmacinnes2391]

As a more full explaination, take a simpler equation in the form y'+h(x).y=g(x) where y'=dy/dx The integrating factor, I(x), is what you multiply the equation by in order to attain a product of differentiation on the left hand side. As we all know, (u(x)v(x))'=u(x)v'(x)+u'(x)v(x) We have y and y' on the LHS, so take u(x) to be y and u'(x) to be y', we now need a v(x) such that v(x)y' + v'(x)y is a product (of differentiation) Without defining I(x), we can say that I(x)y' + I(x)h(x)y = I(x)g(x) So I(x) is out v(x), and I(x)h(x) is our v'(x). Since I(x)=v(x), we can say that I'(x)=v'(x), so I'(x) = I(x)h(x). Continuing, (I'(x)/I(x))=h(x) And since d/dx(ln(f(x)) = f'(x)/f(x) (by the chain rule), we can now integrate both sides to get: ln(I(x))=int[h(x)dx]+c (take c to be 0 as any integrating factor will suffice) So raising e to the power of both sides, I(x)=exp(int[h(x)dx]) You could then solve the original equation for y, but I'll leave it there since I only wanted to deduce the integrating factor

[@GinLottus]

What a beautiful explanation!! It remains me the way Zill's book approach the non-exact DE solution (in the 9th ed. there is a subsection of these topic). Just a little correction in the final step of your answer: the solution is F(x,y), but the DE is equal to dF(x,y) = 0. So, F(x,y) = C. That means, the right answer is F(x,y) = x^2/y + x = C, and the explicit solution is y = x^2/(C - x).

[@slavinojunepri7648]

He kind of stated it at the last minute without writing it down. Your comment makes it clear because the solutions of this deferential equation are in fact the level curves of F(x,y). Well done 👍

[@0.p_tlt935]

I haven watched so many videos, but the way u teach it just made too much sense. Thank you for saving my grade

[@zpocrm]

Thank you, sir. Your video is very organized and provides a detailed explanation of the process for finding the original F(x, y). The formula part can be confusing, but you make it easy to understand and remember.

[@punditgi]

Prime Newtons is the best! 🎉😊

[@debmalyabanerjee1134]

Thank you so much Sir for keeping my request. Please do bring a lot more videos on differential equations in which we can solve homogeneous equations, bernoulli equations and second order differential equations.

[@rickyzhang7424]

best explaination！！！ love from china ❤

[@samarthpriyadarshi257]

Thank u so much sir, I am from india in grade 8 , we generally learn differential calculus and other things in 12 grade but i was easily able to learn new things with your videos. It was really helpful

[@hosseinmortazavi7903]

Hello Mr Newtons you explained the best

[@chemdelic]

You explain things so well!!

[@skwbusaidi]

The function F , we have introduce for exactness. The final answer x^2 / y + x+ C1 = C2 x^2 / y + x = C

[@johnnyblazer513]

Thank you, sir! I was struggling with exact equations stuff, but now I think I am prepared to my exam. Definitely like)

[@PrimeNewtons]

Glad it helped!

[@SonyuyDzesinyuy]

This is what I needed thank you so much sir for clearing my doubts 🙏

[@renzalightning6008]

This reminds me of the Total derivative and Cauchy-Riemann equations (apologies if you mentioned them, I skipped and spanned around the video I will admit but it all seemed familiar despite not practicing it for years)

[@mathnerd5647]

nice video

[@KineHjeldnes]

brilliant explanation! So easy to follow :D

[@vitotozzi1972]

Awesome explain!!! Cool!!!!

[@KingsleyOpoku-i7q]

I'm confused here , shouldnt the intergral of 1/y be lny and not ln(1/y) ,any reasons to his answer at 10:38

[@iliast7198]

It is lny but the -2 moves over and e and ln cancel so it becomes y^-2 which is 1/y^2

[@chestertamayo1894]

what happened to the absolute value?

[@slavinojunepri7648]

Does a solution exist in case there's no integrating factor being a function of x only or y only?

[@lawrencejelsma8118]

I would not have thought of that by some My and Nx and M and N formulation. I was trying to form some u(x,y) substitution of u(x,y) = y/x and then the y' = u^2 + 2u thought process. Partial of u(x,y) respect to and y and x is then = -y/(x^2) + y'/x chain rule sort of thing in terms of u'(x,y) ? Then getting lost in manipulations further knowing about trying to keep it more dy/dx and y/x related. 🤔

[@barryzeeberg3672]

3:07 the equivalent equation could have been written differently, where everything is divided by x^2 (for example). Then M and N would be different than you show here.

[@gillestondo341]

Magnificient!

[@vasilisr7]

Nice video

[@holyshit922]

There we have two other and simpler methods 1. Homogeneous ODE 2. Bernoulli ODE so why exact ? Integrating factor for homogeneous equation mu(x,y) = 1/(xM(x,y)+yN(x,y)) Integrating factor for Bernoulli equation mu(x,y) = exp((1-r)Int(p(x),x))y^{-r} Warning Bernoulli equation must be in the form y'+p(x)y=q(x)y^{r} If we find two independent integrating factors we can simply divide them to get solution In fact we can get two independent integrating factors for this equation based on information that this equation is both homogeneous and Bernoulli Checking condition for exact equation after multiplying equation by integrating factor is always a good idea because we check correctness of calculated integrating factor

[@2R2Y2X2]

ty c u next final :D

[@manu-mm4pc]

Also what happin when both steps are true? at 7:04

[@cloudx4392]

watching one day before exams💉

[@junenaya2736]

What happened to the -2 at 10:38 ??

[@junenaya2736]

Oohh i got it Its a property of ln (a)lnx = lnx^a

[@habeebkinlolu9668]

❤

[@manu-mm4pc]

Hello I dont get how is ln 1/y^2 and not -2ln 1/y at 10:33

[@dennjerrr]

Same

[@martialfestus3352]

@@dennjerrr He applied the rules of logarithm which says -2ln(y) can be rewritten as ln(y)^-2 = ln(1/y^2)

[@kragiharp]

Oh my, I have forgotten so much. I can't even follow your video. First time this happens to me. Shame on me. 😮 I'll continue anyways. Never stop learning. 😊

[@張茗茗-y9i]

It's also homogeneous.

[@DEYGAMEDU]

sir I have sent you a question in your mail. please solve that sir.

[@rohamyaghoubisabet1650]

The worst possible solution