Separable Differential Equations

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Күн бұрын

Пікірлер: 24

@afzalmehdi3269 10 ай бұрын

I like your way of explaining. You talk soft and clear and more over use chalk and black board unlike others using digital pen and don't even talk English properly. Keep it up bro

@adw1z 10 ай бұрын

And now solving first order non-linear differential equations - I have never seen anyone cover as wide a range of questions as you do, and it's a joy to watch. Keep going and never stop learning 🫡

@holyshit922 10 ай бұрын

There are three main types of first order ordinary differential equations 1. Separable ODE 2. Linear ODE 3. Exact ODE Other types of first order ordinary differential equations which can be seen in textbooks can be reduced to this three types by substitution, integrating factor , introducing parameter

@johnconrardy8486 3 ай бұрын

your fun to watch great teacher

@Private-se1gl 10 ай бұрын

very good بسیار عالی حل کردید 🎉

@michaelbaum6796 10 ай бұрын

Very nice example, thanks👍

@richardbraakman7469 10 ай бұрын

Those trig functions keep showing up where they weren't invited :)

@noblesleem1077 10 ай бұрын

😂

@channelbuattv 10 ай бұрын

exponentials are even worse. you'll see lots of them in diff eq

@abhishankpaul 6 ай бұрын

Logarithms and exponents have a different opinion in this matter

@surajsk7315 10 ай бұрын

solve Jee Advanced problems of Mathematics

@Spider70465 10 ай бұрын

Amazing 🤩

@ChimezieFredAnaekwe 10 ай бұрын

❤

@antonionavarro1000 9 ай бұрын

Another solution: f(t)=(1-t)/(1+t) If you check it will be correct (with t not equals to -1).

@MASHabibi-d2d 10 ай бұрын

Thanks for an other video

@holyshit922 10 ай бұрын

For future video Let T_{n}(x) = cos(n*arccos(x)) Let t = arccos(x) y(t) = cos(n*t) y'(t) = -n*sin(nt) y''(t) = -n^2cos(nt) y''(t) = -n^2y(t) y''(t) + n^2y(t) = 0 Lets change of independent variable t = arccos(x) dt/dx= -1/sqrt(1-x^2) dy/dt = dy/dx * dx/dt dy/dt = dy/dx * (-sqrt(1-x^2)) d^2y/dt^2 = d/dt(dy/dt) d^2y/dt^2 = d/dx(dy/dx*dx/dt)*dx/dt d^2y/dt^2 = d/dx(dy/dx * (-sqrt(1-x^2)))*(-sqrt(1-x^2)) d^2y/dt^2 = sqrt(1-x^2) * d/dx(dy/dx * sqrt(1-x^2)) d^2y/dt^2 = sqrt(1-x^2) * (d^2y/dx^2 * sqrt(1-x^2) + dy/dx * (-x)/sqrt(1-x^2)) d^2y/dt^2 = (1-x^2)d^2y/dx^2 - xdy/dx y''(t) + n^2y(t) = 0 d^2y/dt^2 = (1-x^2)d^2y/dx^2 - xdy/dx + n^2y(x) = 0 y(1) = 1 Solve equation above with power series

@holyshit922 10 ай бұрын

Problem is that i get factorial of negative number or division by zero while expanding fraction to factorial and later to binomial coefficient

@nicolascamargo8339 10 ай бұрын

Genial

@aaronmisquith9341 10 ай бұрын

Why is it that when you took the integral of dy/y²+1, you didnt add a +c onto it like you did for 1/t²+1?

@carultch 10 ай бұрын

I'm assuming you are talking about time stamp 3:36. There is a constant of integration in both integrals, so technically, you can have a +C1 on the left integral, and a +C2 on the right, as you are suggesting, which gives us: arctanh(y) + C1 = -arctanh(t) + C2 However, we also can see that these two integration constants are not independent of each other. We can subtract C1 from both sides and get: arctanh(y) = -arctanh(t) + C2 - C1 Since it doesn't matter how we set C2 and C1 relative to each other, we can just combine them to one constant of integration, and get: arctanh(y) = -arctanh(t) + C Because this step happens in separable differential equations all the time, it is common to just keep it simple, and only add a +C on one of the integrals, but not the other.

@carultch 10 ай бұрын

Generally, you will only have an undetermined constant in the final general solution, for every order of differentiation involved in the highest derivative. This is how you can anticipate how many of the constants of integration to either absorb each other in intermediate steps, or ultimately cancel through other algebra as you post-process your integration results.

@himanshuhooda8762 10 ай бұрын

Eary for iit jee students

@honestadministrator 9 ай бұрын

f ' ( t) /[ (f( t)) ^2 + 1] + 1/( t^2 + 1) = 0 d ( arc tan ( f(t)) + arc tan (t) ] = 0 arc tan ( f(t)) + arc tan (t) = arc tan ( 2) + arc tan ( 3) = π/2 - arc tan (1/2) + π/2 - arc tan (1/3) = π - arc tan ((1/2 + 1/3) / (1 - 1/6)) = π - arc tan ( 1) = 3 π /4 Hereby f (t) = tan ( 3 π /4 -arc tan ( t)) = (1 - t) /( 1 + t)