Solve a differential equation by substitution: dy/dx=(x-y+5)^2 (answer on WolframAlpha)

[@abdraoufabusoua8004]

Instead of long division, you can add 1 and substract 1 . So you reach to the same result faster .

[@pleindespoir]

I have studied maths at more than one university and was reading (and working severly) tons of books about differential equations but it lasted nearly thirty years to find someone who is able to explain these things as clearly as you are doing. I appreciate your vids so much! I wished more professors would take their time to listen your explanations but in staying to confuse their students with their bluring explanations that everybody will be lost in very dense fog.

[@blackpenredpen]

Thank you! It is my pleasure to have you here.

[@z01t4n]

You could look at the missing solution, y= x + 4 as being lim( y = 2/(1+Ce²ˣ) + x + 4 ) when C→±∞; so just write y=2/(1+Ce²ˣ)+x+4 C ∈ ℝ∪{+∞}∪{-∞} and let your teacher struggle to explain you why infinity is not a number and why it cannot be part of any domain. :-))

[@DRACOBUCIO]

lol

[@deeptochatterjee532]

z01t4n Woah man how'd you get all those math symbols on here

[@Koisheep]

Actually in topology we have like an extension of the standard real like which includes the ∞s called extended real line. Basically you define the point ∞ as the limit of any divergent sequence. (It is basically the circumference btw)

[@6612770]

Is it valid to obtain The Missing Solution by taking Result = Limit as C goes to Infinity?

[@jarikosonen4079]

That's maybe the safer way?

[@overlordprincekhan]

I also thought so

[@krukowstudios3686]

2:36 can’t you just integrate it right away? Integral of 1/(1-v^2) dv is just tanh^-1(x), and on the right you get x?

[@cbbuntz]

5:26 ln(1+v) - ln(1-v) = 2x+C1 is just 2atanh(v) = 2x+C1. I feel like that property could have reduced the amount of algebraic manipulation needed. atanh is cool because dealing with integrals involving products with logs of other functions is scale invariant. You can solve for the exponent c with stuff in the form a*(1-x^2)^b*((1+x)/(1-x))^c and it doesn't matter what the other indeterminants are are. I works for a wide range of functions in a similar form. Take something kind of nasty looking like a*(1+x^2)^b*(((1+x)^(1/2)+x)/((1+x)^(1/2)-x))^c and the beauty is all the extra stuff cancels in the log domain and it reduces to atanh. It's basically just logistic regression shifted to a domain of -1 to 1

[@deangel9128]

And coth^(-1) x would be the rest of the domain (-inf,-1)U(1,inf) if I'm not mistaken.

[@overlordprincekhan]

6:18 I believe he can do the same technique as of in componendo-dividendo

[@davidbrisbane7206]

If instead of the given problem, if we solve dy/dx = (x + y + 5)², then we find the solution is y = tan(x + c) - x - 5. Now if 1 + v² = 0 ⇒ v = ±i. Now as i ∉ ℝ, so there are no missing solutions.

[@technicbrasil]

Can we use the inverse hyperbolic tangent for the integration step or we need to use the partial fraction technique ?

[@blackpenredpen]

euardo balint depends on if ur teacher accepts that result or not. I understand they are equivalent so it would be an okay for my class

[@6612770]

Would you please expand on your point at 16:10 as to why doing such a division can lead to 'loss of answers' ?

[@AndDiracisHisProphet]

When you divide you could "divide by zero"

[@yaeldillies]

Always think to check if you do not divide by zero, or make a special case of that

[@TaylorNguyen2909]

This is why in my ODE class whenever we divide separable equations we have to check the case of 0 first, since most cases its the og solution

[@franciscoespinosperez4544]

The missing solution can be obtained by taking the limit when c goes to infinity.

[@cameronspalding9792]

By Picard's theorem if y(a)=b we have a unique solution

[@akuviljanen4695]

Can you do dy/dx=cos(x+y)? I have thought about it for a long time and I think I managed to do it with this video, but it would be great to see your solution as well. Thanks for awesome videos :)

[@meh7272]

You did by inputting x+y = v ?

[@arpitdas4263]

Perform a substitution and convert into a linear form

[@alexmarkov9918]

Expression for v can be expressed as tanh(x+c). Looks better IMHO. :-)

[@arshdeep011]

eng. speaking v kra dena ..good job😉😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊

[@pompei2]

May let C=infinity, and we catch y=x+4

[@habiballahi7579]

Grate job

[@tuancong4299]

Perfect

[@BriceLavorel]

It's old but you made a mistake (which is not one at the end), your final constant C cannot be 0 because C1 is a real value, so C2 is strictly positive (0 is forbidden), so C3 is real without 0. Finally, in your family of solution, the case C=0 is not part of it. It is not a mistake at the end because you treated the case of Y=X+6, considering it as a non missing solution (but it is a missing solution).

[@suratbisokarma1171]

dy/dx=(x+y+1)2 how to you solve

[@tahaabujrad7806]

Let c gets bigger and bigger until the whole term becomes zero .. How y=x+6 is not a missing solution ...you said that c must equal to zero ...and if we back to what does c equals we find that it is e^c1 so e^c1 =0 and this can not happen only If you let c1= -inf ..

[@blackpenredpen]

I explained that at 17:30, take a look : ) c=0 =>2/(0+1)+x+4 = x+6

[@tahaabujrad7806]

blackpenredpen ..first thank you for replying..second how can C equal to zero ... I mean the constant C before being C it was equal to -+e^C1 . So if we keep it as it is it can not equal to zero ..in other words the the domain for C is R - {0} .

[@tahaabujrad7806]

My point is at the time from 5:50 to 6:00 ..where you said " c1 is a constant e is a constant e^c1 is another constant (but can not equal to zero) " then you write it as c3 ..and you allowed it to accept zero as a value ...(how come) ..let's assume that you kept it as e^c1 .how will you get the ..particular solution (x+6) ..please reply again .

[@MarcoMate87]

It's true. By definition, c_3=e^(c_1) or c_3=-e^(c_2), so it can be any real number, positive or negative, except 0. But we can simply extend the definition of c_3, allowing it to be also 0, and we recover one of the missing solutions. The two missing solutions are simply constant solutions of the separating variable equation for v. In any equation of the form v'(x)=f(x)*g(v), if exists a number v_0 so that g(v_0)=0, then v(x)=v_0 is always a (constant) solution of the equation.

[@tahaabujrad7806]

MarcoMate87 thanks ..one more thing I will assume that x+6 and x+4 are both missing solutions. In the video he proved that x+6 is not missing by letting C =0 which implies that c1 in e^(2x+c1) to be equal to -inf .(he used inf) by the same way I will prove that x+4 is not s mission solution by letting C=+inf >>c1 in e^(2x+c1) to equal to +inf (I used inf to prove it same as he did(by indirect way))

[@montsaintleondr7491]

Why the goodness Wolfram? Why not using more advanced software?

[@blackpenredpen]

MontSaintLeon Dr such as?

[@gashawbezatadiyos3786]

The answer is not correct because in the solution ln(1+v) + ln(1-v) not negative sign between them

[@ayushranjan6807]

Nope. integral dx/(ax+b) = (ln|ax+b|)/a. In integral du/(1-u), a=-1. So you divide (ln|1-u|) by -1, which is -ln|1-u|

[@mamtagoel1242]

Other solution is y=x+5-tanh-1(x+c)

[@overlordprincekhan]

12:39 Is that an Iphone?

[@john-athancrow4169]

#FactorTheTerms (CamelCase)(Hashtags(#):1)