Came to this channel for integrals stayed for differential equations (I love them)

"sorry, i just turned jamaican for a moment" ... 😂

If A is a free constant of integration then no need to carry 1/sqrt(A) all the way through isn't it so? Also think you answer misses a ± somewhere (the absolute values of the logs which you didn't carry and there's a spot where you have 1/2 on the RHS which could have been transposed into a 2 on the LHS and become a square in the log expression.

This was a great example of how to solve a nonlinear autonomous ODE. Well done!

Hi, "ok, cool" : 0:19 , 7:58 , "terribly sorry about that" : 4:42 , 5:01 , 6:12 .

Hi bro, could you make a video proving all the Maxwell equations?, I know there a lot of them, but I like how you explain, great video

Alfred Steele: I got Ax + B = sqrt( 1- y^2) + arcsin(y). Begin with his Ax + B = \int( sqrt( (1-y)/(1+y) ) ) multiply top and bottom by sqrt(1 - y) and get the integral \int( (1-y) /sqrt(1-y^2) ) This splits into two easy integrals \int( 1/(1-y^2) ) - \int( y/(1-y^2) ) = sqrt( 1- y^2) ) + arcsin(y)

Fantastic

What would happen if I would integrate 1/(1-y^2) as arctanh(y) ?

Can I write the answer as following? x=A(√(1-y²)-cos⁻¹y)+B

xe^c+C1=-2arctg((1-y)/(1+y))+√(1-y^2)

1:42 if you wanna be real 🤓 about it, C²≠1

Bombooooclaaaat

That was an annoying one for me. Pretty difficult.

Jamaican vampire?

delicious

Adding another horrible accent to the collection smh

When I solve these problems I'd like to be extra rigorous and very careful. Goal: Find all y: 𝔻→ℝ such that for all x in 𝔻 ; y“(x) = y’(x)^2/(1−y(x)^2) First, note the restrictions caused by the equation: 1−y(x)^2 appears in the denominator so it should be nonzero for all x in 𝔻 ie. y(x) ≠ ±1 for all x in 𝔻 Next step divide both sides by y’(x) but this assumes that y’(x) ≠ 0 so we have to exclude the values that set y’(x) to 0. We will call their set S the equation becomes for all x in 𝔻\S ; y“(x)/y’(x) = y’(x)/(1−y(x)^2) Now we will only consider solving over an interval in 𝔻\S and here is why: The derivative of the piecewise function f : ℝ\{x0}→ℝ such that f(x) = a if x>x0 and f(x) = b if x

@YouTube_username_not_found

5 күн бұрын

I realized I made a mistake. For the case −1