Solving A Differential Equation | Two Methods

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[@zahari20]

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.

[@adandap]

Yes, for sure. It falls out in half a dozen lines when you do that.

[@marciliocarneiro]

really.. you´re right..

[@duaafatima-nk3bo]

nice contribution for this era people

[@IantoCannon]

That first substitution was filthy

[@Shadi677]

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

[@ismaelcastillo188]

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

[@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

[@kingoreo7050]

Thank you for making more calculus and trigonometry videos!

[@krunoslavregvar477]

Nice as well as educational. 👍✌️👍

[@RaviBRUH]

you can directly use byparts in the question

[@user-dy5pi7bm6f]

Great video! More DEs would be great

[@yoav613]

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.

[@phild8095]

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

[@SyberMath]

I don't like stats. No offense

[@angelmendez-rivera351]

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(arsinh) = (-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.

[@joyli9893]

WOW! Integral dy is really equal to y!

[@giuseppemalaguti435]

y=(1/2)(xsqrt(1+x^2)+arcshx)+c

[@jimmykitty]

Awesome ♥️

[@SyberMath]

Thanks 🤗

[@SyberMath]

Long time, no see! 😊

[@behrensf84]

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

[@SyberMath]

😲😄

[@93techie]

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

[@adamfiser7645]

Yes, there (9:10) should be either +ln(sqrt(x^2+1)+x) or -ln(sqrt(x^2+1)-x), because t^(-1) = sqrt(x^2+1)+x. He confused t and t^(-1).

[@f1sk313]

Там действительно ошибка

[@duckymomo7935]

Trig substitution is the best choice here

[@anestismoutafidis4575]

=> Int (X^2•dx-dy)/2

[@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.

[@del66404]

👏👏👏👏👏👏👏

[@shafin3365]

👍

[@Mephisto707]

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

[@usdescartes]

It WOULD be, but what we are really doing is using the definitions via the sides of a triangle... which, as you point out, are always positive. But don't forget that x is inside a square root (which is always positive itself), and when we do a back-substitution it still will be. It (almost) never happens that a situation arises where it would matter, so just like if our final answer had a ln|x| in it but we knew x was always positive, we'd write it ln x and get rid of the absolute value bars, we ignore the |sec x| and just cal it sec x because we MUST be in a region where that square root will always be positive anyway. I imagine it DOES get tricky if we are doing complex analysis, but my last complex analysis class was 30 years ago, and I don't remember if/how we handled trig sub for those problems... if we even did any.

[@manjoker]

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

[@fartsniffa8043]

its not an elementary integral.

[@scottleung9587]

I just cheated by looking at the integral table!

[@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

[@gabrijel5341]

You are indeed correct, dy/dx is a symbol as a whole, but in this case you can treat it as division and it will not matter and it helps solving this equation easier.

[@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.

[@awindwaker4130]

Diophantine equations are only interested in integer solutions.

[@gdtargetvn2418]

what the

[@imonkalyanbarua]

@@awindwaker4130 thanks. I realised it later. 😊👍

[@alextang4688]

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

[@jayshreeram.374]

This is not very fast question

[@AbouTaim-Lille]

This is an elementary problem. First thing it is of rank 1 unlike the Bessel differential equation for example . And 2nd thing the differential y' is solvable as a function of x . And after all we are still talking about ordinary differential equations !! ..... So it can be integrated directly .

[@mcwulf25]

Is that t substitution the same as tan(theta/2) substitution? When I see √(1+x^2) my inclination would be to go straight in with tan(theta) but I see someone has shown how that gives us sec^3(theta) which is a nuisance to integrate.

[@timeonly1401]

Let J = integral{ [sec(x)]^3} dx = integral{sec(x) [sec(x)]^2} dx . Let u=sec(x) & dv=[sec(x)]^2. So, du=sec(x)tan(x)dx & v=tan(x). J = sec(x)tan(x) - integral{sec(x) [tan(x)]^2} dx J = sec(x)tan(x) - integral{sec(x) ([sec(x)]^2 - 1)} dx J = sec(x)tan(x) - integral{[sec(x)]^3} dx + integral{sec(x)} dx J = sec(x)tan(x) - J + ln |sec(x) + tan(x)| + C Solving for J: J = (1/2) [ tan(x)sec(x) - J + ln | tan(x) + sec(x) | ] + C Then back-sub to finish.

[@timeonly1401]

In the last line of equation (right after "Solving for J"), there obviously shouldn't be a J term on the RHS. :-(

[@mcwulf25]

@@timeonly1401 Yes because it has gone to the left 👍

[@doyourealiseiamaghost5862]

sub x with sinh(u) for easier approach

[@roldansosabarajas1213]

Podría hablar un poco más lento por favor

[@arnabbose5398]

t or not to t

[@Vladimir_Pavlov]

I give you the third way. Integration "in parts". Int = ∫√(x^2+1)dx = | u=√(x^2+1) ⇒ du=x/√(x^2+1) , dv=dx ⇒v=x | = x*√(x^2+1)- ∫x^2dx/√(x^2+1) =x√(x^2+1)- Int +∫dx/√(x^2+1). | the last integral is considered tabular all over the world - "long logarithm"| 2* Int= x√(x^2+1)+ ln(√(x^2+1) +x)+C, Int= (1/2) *x√(x^2+1)+ (1/2)*ln[√(x^2+1) +x]+C, the sign of the absolute value of the logarithm argument does not need to be set. Positivity is fulfilled .)

[@broytingaravsol]

y=c±[x√(1+x²)+ln|x+√(1+x²)|]/2

[@barakathaider6333]

👍