Solving A Differential Equation | Two Methods

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

If you need to post a picture of your solution or idea:
#CalculusProblems #DifferentialEquations #Integrals
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :

[zahari]

The best substitution for the integral is x = sinh(t). Then you get integral of cosh(t)^2 where cosh(t)^2 = (1/2)(1+cosh(2t)) etc.

[marciliocarneiro]

I used the transformation x=tan(u) and obtained a shorter resolution because I obtained the integral of sec(u)^3 which is less complicated

[Shadi]

The tan approach is amazing .. so we can replace the "ln" at the end with arc sinh(x)

[ismaelcastillo]

In this kind of integrals is usually clever to use t=sinhx as cosh^2-sinh^2=1

[angelmendez-rivera]

Let y : U —> R such that y is continuously differentiable everywhere, and such that D(y)(x)^2 – 1 = x^2 everywhere. This is equivalent to D(y)(x) = (–1)^m•sqrt(x^2 + 1) everywhere, with m = 0 or m = 1. Let g[m](x) = (–1)^m•sqrt(x^2 + 1) everywhere. Now, D(y)(x) = g[m](x) everywhere, with m = 0 or m = 1. Here, we can apply the Riemann integral on [0, t] to both sides of the equation, and by the fundamental theorem of calculus, the Riemann integral of D(y) is equal to y(t) – y(0). What remains is for us to find the Riemann integral g[m].

Notice that sqrt(x^2 + 1) = sqrt(sinh(arsinh(x))^2 + 1) = sqrt(cosh(arsinh(x))^2) = |cosh(arsinh(x))| = cosh(arsinh(x)) = cosh(arsinh(x))^2•1/sqrt(x^2 + 1) everywhere. This is important, because if h(x) = 1/sqrt(x^2 + 1) everywhere, then D(arsinh) = h. Also, notice that if j(x) = x/2 + sinh(2•x)/4 = x/2 + sinh(x)•cosh(x)/2 everywhere, then D(j)(x) = 1/2 + cosh(x)^2/2 + sinh(x)^2/2 = 1/2 + cosh(x)^2 – cosh(x)^2/2 + sinh(x)^2/2 = 1/2 + cosh(x)^2 – 1/2 = cosh(x)^2. As such, we can write that g[m] = = (–1)^m•D(j°arsinh) = D((–1)^m•(j°arsinh)), so the Riemann integral must be equal to (–1)^m•j(arsinh(t)) – (–1)^m•j(arsinh(0)). Thus, we have y(t) – y(0) = (–1)^m•(j(arsinh(t)) – j(arsinh(0))) = (–1)^m•j(arsinh(t)) everywhere. j(arsinh(t)) = arsinh(t)/2 + t/2•sqrt(t^2 + 1), hence y(t) – y(0) = (–1)^m/2•(arsinh(t) + t•sqrt(t^2 + 1)) everywhere.

[phild]

Did all this 45 years ago, worked in process engineering and never used it. Advanced Stats would have been more of a help.

[techie]

When you replace ln|t|, why did you use t = sqrt(x^2+1)+x? shouldn't it be sqrt(x^2+1)-x?

[RaviBRUH]

you can directly use byparts in the question

[kingoreo]

Thank you for making more calculus and trigonometry videos!

[yoav]

Very nice! It can be solved also by IBP (x)' sqrt(1+x^2). Another nice substitution for this integral x=i sinu and you can use the identity i arcsin(-ix)=arcsinhx.

[Mephisto]

10:30 Isn't the square root of (sec x)^2 absolute value of (sec x)?

[behrensf]

Darn… it’s been so long I don’t remember how to do any of these… I just use a spreadsheet now…

[joyli]

WOW! Integral dy is really equal to y!

[manjoker]

please solve this is integral ln(cscx+cscx cotx) dx

[imonkalyanbarua]

Fantastic solution! I was just wondering, what if we bring x^2 to the LHS and take 1 to the RHS. Then we get a form:

a^2 - b^2 = 1

Can't we then solve it like a Diophantine equation? It will be easy although the answers will be different of course, but is it wrong? Kindly explain it to me. Thank you.

[dienosorpo]

But my teacher always told me dy/dx is a symbol as a whole, not a damn division 😭😭😭

I do not know anything about calculus but I feel like that's incorrect.

The integral has a dx at the end to make it clear that you are integrating with respect to x.

And the integral with respect to x of the derivative with respect to x is the function.

That's why on the other side there appears a "dx" because you integrated both sides with respect of x. Not because you multiplied both sides by dx.

Am I correct? Or not? I'm just in highschool so I could be wrong

[mosquitobight]

I started it wrong because I misread the square of the derivative as the second derivative. I wouldn't mind seeing the solution to that, btw.

[alextang]

This is a standard problem in advanced level mathematics at high school. 😋😋😋😋😋😋

[dypibmf]

Great video!
More DEs would be great

[scottleung]

I just cheated by looking at the integral table!