solve differential equation with substitution

Рет қаралды 138,141

Күн бұрын

Пікірлер: 65

@VetteTTV12 5 жыл бұрын

Seriously wish you taught at my university. All of our higher class, calc3 and up, have really crappy instructors that care more about theory and proofs than actually showing us how to solve the problem. So a big thank you for helping those of us that need someone like you to break it down quick and simple in order to understand. If you have a patreon page or something let me know, you deserve some form of compensation for this. If it wasn't for you I would've failed a few different tests.

@79Shotinthedark 6 жыл бұрын

This is my first time watching one of your videos. I appreciate how you take your time with the problem and that you write very clearly (surprisingly hard to find). Using the two colors made it easier to follow. I learn a lot of my math from KZbin and this was very helpful. Thank you.

@sam-kx3ty 4 жыл бұрын

You’re one of the best math lecturers in the world please keep it up .

@HonsHon 4 жыл бұрын

Thank you! Helping me so much in preparing for the final in my DE class. Ever since I was in Calc 2 I have been watching these, and they are so helpful.

@chuaprincecarl9845 4 жыл бұрын

the marker switch is smooth af, Michael Jackson is proud.

@varunnayak5369 6 ай бұрын

criminal 😂

@Kapomafioso 7 жыл бұрын

4:55 you could in fact solve for v. Let's say we have it in some final form like: sec(v) + tan(v) = r (r stands for whatever it is on the right-hand side) Then we do this: 1+sin(v) = cos (v) r 1 + (e^iv - e^-iv)/2i = (r/2) (e^iv + e^-iv) Now let e^iv be p. Then we have: 1 + (p + 1/p)/2i = (r/2) (p + 1/p) //multiply by 2 i p 2 i p + p^2 + 1 = r i p^2 + r i This is some polynomial in variable p, solve for p, make out a logarithm out of it and "see" the arctan function in it. Other approach is: rewrite sin(v) - r cos(v) as: sqrt(1+r^2) sin(x + arctan(r)) and you will finally obtain the same arctan formula. Of course, there would be some decisions like which root to take and add + 2*pi*integer somewhere when taking inverse functions, but, youknow, some people say: a differential equation is not complete unless you provide a sufficient set of initial and/or boundary conditions ;) so after you clarify initial condition, there should be no arbitrarity.

@ChefSalad 6 жыл бұрын

Without WolframAlpha, I know how to solve for v. Start by taking the exponential function of both sides, and relabeling the c: sec(v)+tan(v)=C₁*e^(-1/x). Change sec and tan to sin and cos and combine fractions: (1+sin(v))/cos(v)=C₁e^(1/x). Shift the sin and cos to cos and sin: (1+cos(v+π/2))/sin(v+π/2)=C₁e^(−1/x). Reciprocate: sin(v+π/2)/(1+cos(v+π/2))=C₂e^(1/x). Use the tangent half-angle identity: tan(v/2+π/4)=C₂e^(1/x). Take the arctan of both sides. v/2+π/4=arctan(C₂e^(1/x)). Solve for v: v=2arctan(C₂e^(1/x))−π/2. Substitute back in v=y/x²: y/x²=2arctan(C₂e^(1/x))−π/2. Solve for y: y=2x²arctan(C₂e^(1/x))−πx²/2. BAM! Solved for y.

@brandindia7672 7 ай бұрын

🤔

@ageofkz 7 жыл бұрын

Is there a special name for this sort of functions where you make a substitution to solve it? For example, homogenous 1st ODE you will substitute f(y/x)=f(v), v=y/x.

@DougCube 7 жыл бұрын

Here is the closed-form solution: 2(x^2)arctan(tanh((Cx-1)/(2x))). Not that anyone cares...

@AkshayMuraliNerd098 6 жыл бұрын

DougCube how did u get that

@srpenguinbr 6 жыл бұрын

@@AkshayMuraliNerd098 if you express sec(x) and tan(x) in terms of sin or cos, you can isolate the y

@ivypellerin3166 3 жыл бұрын

Thank you for showing how we get the substitution for dy/dx my profs like to skip intermediate steps also loved the flawless marker flipping hahaha

@JesusGarcia-ox3jj 7 жыл бұрын

you should do more of these and Bernoulli's equation

@blackpenredpen 7 жыл бұрын

Jesus Garcia they r coming this weekend

@Samir-zb3xk 6 ай бұрын

instead of using ln|sec(v)+tan(v)| for integral of sec(v) if we use artanh(sin(v)) (which is equivalent to ln|sec(v)+tan(v)), it makes it easy to solve for y i got y=x²arcsin(tanh(-1/x+c)) as final answer

@bioengboi137 3 жыл бұрын

“There’s no way to isolate the v”. Well even though int{sec z dz} = ln|sec z + tan z|, I think the more proper form to rewrite it with one input & no absolute value; +/-, is that the int{sec z dz} = arctanh(sin z). So with some “function sliding”, y = x^2 arcsin(tanh(c - 1/x)). Cool fun fact the derivative of arcsin(tanh z) = sech z

@williamadams137 5 жыл бұрын

Could the answer be sec(y/x^2)+ tan(y/x^2) = Ce^(-1/x) ? Note : after removing the absolute value, i put plus or minus on the other side and a “plus or minus” constant is another constant.

@candlelightc4699 Жыл бұрын

its 5 years later but thank you for the very clear explanation

@SuperKSA707 2 жыл бұрын

Thank you! You're really helping us. god bless you

@shaunakmehal6412 23 күн бұрын

Beautiful

@BriceLavorel 2 жыл бұрын

if you use arctanh(sin(x)) as a primitive of 1/cos(x), everything become easier : y = x^2 * arcsin(tanh(C-1/x))

@cormackjackson9442 Жыл бұрын

Which type of ODE is this?

@SaifUlIslam-lw3dm 11 күн бұрын

It's 7 years later... And still thanks...

@someone2879 5 жыл бұрын

This is really helpful ... Thanks for uploading!

@nathangething6418 2 жыл бұрын

thank u this was very helpful

@aditmistry4936 5 жыл бұрын

How the hell do you switch markers so fast!?

@Jjdumott 4 жыл бұрын

it is because they're in the same hand same time i was wondering that too lmao

@ramakrishna-bi1co 6 жыл бұрын

Well explained

@dalek1099 4 жыл бұрын

Sec(u)+tan(u)=tan(1/2u+pi/4). Ln|tan(1/2u+pi/4)|=ln(tan(+-1/2u+pi/4))= -1/x+C tan(+-1/2u+pi/4)=Aexp(-1/x) +-1/2u+pi/4=arctan(Aexp(-1/x))+mpi. u=2arctan(Aexp(-1/x))+(2n-1)pi/2 y=x^2[2arctan(Aexp(-1/x))+(2n-1)pi/2]

@CardThrower-rb6eg 4 ай бұрын

man thanks for saving me for my exam later

@MothyEmms Жыл бұрын

gotta love an asian math teacher!

@nra-sheta1882 Жыл бұрын

He is a living legend 🎉

@willlesslie6825 3 жыл бұрын

Getting some Doctor Who, the Ood Vibes here lol

@spiritgoldmember7528 6 жыл бұрын

You can solve for v using a Weierstrass substitution.

@shex9002 7 ай бұрын

BRO THIS GUY IS THE BEST

@ipekisgin1608 5 жыл бұрын

Thank you!!!

@samuelminea5520 7 жыл бұрын

the best teacher!

@blackpenredpen 7 жыл бұрын

Samuel Minea thanks!

@YHWHsam Жыл бұрын

five years later and ur still helping! 😂

@TheEdthekidrePvP Жыл бұрын

thank u sir. i love u sir

@alexandermorozov2248 Жыл бұрын

Как выразить y(x) в явном виде? ~~~ How to express y(x) explicitly?

@Samir-zb3xk 6 ай бұрын

instead of using ln|sec(v)+tan(v)| for integral of sec(v) if we use artanh(sin(v)) (which is equivalent to ln|sec(v)+tan(v)|), it makes it easy to solve for y i got y=x²arcsin(tanh(-1/x+c)) as final answer