How to Solve Bernoulli Differential Equations (Differential Equations 23)

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I agree, but you could write this as a constant lol

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y'all call it calc 4, differential equations

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Please see my Chanel (Abdulhussein Alsultani) to see the differences between my method in solving th e examples) Thank you very much.

[@rudnam]

I'm so so thankful for this really. My prof introduced diff eqs, its properties, separation of variables, linear DEs and integrating factors, and bernoulli's equation all within a span of 40 minutes and i could barely understand anything. You're a lifesaver

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[@jarvisty3640]

This link kzbin.info/www/bejne/poSsooikha6ehpY is also a nice video on Bernoulli's Equation.

[@olivermogga6956]

This is the most effective lecture on Bernoulli Differential Equations I have ever had.

[@lindseytreweek8771]

Today I did the second example by myself when a few days I had no clue what any of this was including a foggy idea on u-sub. It's crazy how many leaps in knowledge I'm experiencing from just following these videos. Thank you thank you thank you. Can never be overstated how much your help guides people like me who thought they were hopeless. You're the best!

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I scored pretty good marks in maths in my initial semesters because of him... when he had stopped making vedios i had no choice so left maths and took another subject as minor.. my syllabus for this sem was differential equations even though I dont have the subject anymore I still watch his vedios 😁😁

[@jarvisty3640]

This link kzbin.info/www/bejne/poSsooikha6ehpY is also a nice video on Bernoulli's Equation.

[@daltoncarlson2402]

Yes..., yes you make sense!!!! The hand-waving is yet explained again!!! Thank you, Sir!!! This beats the heck out of a zoom white board where all we get is a non-rigorous proof (because coming from Calc III Students know what that is....not at all), of Bernoulli's DE's, then thrown into examples where the product rule is bypassed when we learned it, LIKE, yesterday. As a student, I do not need five shorthand examples in chicken-scratch on a digital whiteboard over zoom, I need 1 or 2 examples that dive into the details as you do in your videos. You're the Clark Kent of Mathematics, yet be the one who has found the Kryptonite of college paid professors that lack a basic understanding of how to actually teach what they understand themselves!

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For this presentation he's wearing a long-sleeved shirt without the sleeves rolled up. Good idea - this is serious subject matter, and we students can't be distracted by those biceps.

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[@mpmndza]

Just a thought for the first problem at (6:24). If v = y^(-2), then dv/dx = -2y^(-3)dy/dx. If you divide the whole equation by y^3 you will be able to substitute in (-1/2)(dv/dx) directly. From there you can put it in linear form and solve. Saves you from all of the fraction exponents. Hard to explain in text but you essentially do two substitutions that gets you to dv/dx +2v= -2 in less steps. That's actually the brilliant thing about using substitution methods.

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Even after I'm done with the diff eq class this semester, I will anxiously wait for the rest of your lecture videos. Hands down the best explanation anyone has given on Bernoulli's.

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"leveraging the derivative of the power rule to undo a power rule inside of a product rule to undo that to find your original equation"= this video.

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Just watched the whole thing and I finally understood Bernoulli!, thank you so much superman!

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[@TalkswithJinesh]

Thank you professor for the video. Looks like there is a typo at 24:05, Ce^(-2x) - 1, but you wrote + 1

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He rectifies it later at 24:54

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[@jarvisty3640]

This link kzbin.info/www/bejne/poSsooikha6ehpY is also a nice video on Bernoulli's Equation.

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[@jarvisty3640]

This link kzbin.info/www/bejne/poSsooikha6ehpY is also a nice video on Bernoulli's Equation.

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[@abdellatifbourhim8721]

These computations can be simplified. Indeed, set v=y^{-2}, you get that dv/dx=-2y^{-3}dy/dx. Then divide both sides of the equation dy/dx - y =y^3 by y^3. You get y^{-3} dy/dx -y^{-2}=1. So, (-1/2)dv/dx-v=1 or dv/dx+2v=-2 or dv/dx=-2v-2. This is a simple separable equation dv/(-2v-2)=dx. Now integrate both sides: (-1/2)ln(|v+1|)=x+c. This implies that ln(|v+1|)=-2x+C. Finally, apply the exponential and solve for v to get v=-1+Ae^{-2x}.

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This link kzbin.info/www/bejne/poSsooikha6ehpY is also a nice video on Bernoulli's Equation.

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My friend recommended this channel to me. It is really helping me study for my first DiffEQ exam. I did notice a POSSIBLE ERROR at 24:20 bottom right-hand corner. V=Ce^(-2x) -1 does not = Ce^(-2x) +1 Great video. Please keep making videos! This is crazy helpful!

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Bernoulli is a name appeared very often in Fluid Mechanics.

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This link kzbin.info/www/bejne/poSsooikha6ehpY is also a nice video on Bernoulli's Equation.

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As an alternative way to derive the substitution, you can divide both sides of the equation by y^n. Then the indicated substitution becomes easier to see.

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I just wanted to mention that at 54:33, where you get the Linear differential equation, it turns out you can also solve it by using seperable equations. You can bring 6xv to the right side and factor 6x out...

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The first example was also a separable equation, -2 goes to the right side, you factor it out to get -2(1+v). Then just divide by 1+v and integrate

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[@VndNvwYvvSvv]

It's good to practice the theory for linears, but like Bernoulli always gives a result, you can shortcut the linear to y = 1/rho • ∫ rho F(x) dx, or in this case v= and back-substitute. Just be careful to apply 1/rho (or 1/mu) to the entire integral including C.

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[@nagozerafrank158]

sir thanks but which method are you using after obtaining the integrating factor when you are about to reach the step of substituting back V

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[@abdur-raghmaanmatthews5408]

awesome work on explaining this

[@jarvisty3640]

This link kzbin.info/www/bejne/poSsooikha6ehpY is also a nice video on Bernoulli's Equation.

[@zaga32]

Just a suggestion why not divide everything by y^n then just substitute your v=y^(1-n) and dv/dx = (1-n)y^(-n) * dy/DC. That way you avoid having to do the multiplication and simplifying at the end. Maybe that can help :).

[@ivankaraivanov6152]

Thanks for the video. However, I want to point out that at 8:47 you wrote y^2 = v^-1 => y = v^(-1/2). Correct me if I am wrong, but I think you should have written y = (+,-)v^(-1/2) instead. So I am not sure if you are allowed to make the following substitutions as you did.

[@abdallababikir9154]

the plus minus gets absorbed by C

[@robertbrandywine]

The thing I can't "see" is why *minus* n? 1-n can be restated as +1-n or -n+1. If we just wanted a difference of 1 we could have used n+1 (or 1+n). Is it because when you differentiate a function of x^n you always get nx^n-1?

[@belissegiovanni]

THANK YOU

[@petrmasek4506]

The first equation can be done by variable separation right?

[@Mark-di6cz]

Thank you so much for the lectures. Do you think it would be easier to get the equation in standard form, then multiply through by the reciprocal of the y^n on the right side. Then substitute, that way you don't have to solve for y in the v=y term. Just take dv/dx and substitute on the left side dy/dx term.

[@angelicafoster4488]

Hello is there a Differential Equations book that you would recommend?? Which one do you use when you teach?

[@vladislavcaptari2131]

Hey! I'm noticing that at 20:47, the bernoulli differential equation we obtained can be solved by separation. Is this correct or do we have to continue solving the DE using an integrating factor?

[@vladislavcaptari2131]

I have tried solving that bernoulli differential using the separation technique and obtained the same result! I answered my own question I guess... Would there be any reason why I can't do separation?