How to solve separable differential equations (6 examples, calculus 2)

[@feuerwelle4562]

3:57 I think you made a little misteak with the notation, because you put sin(0) instead sin(π)

[@realgwenstacy]

it just so happens that sin(0)=sin(pi)

[@weebtuku]

I think it's just sin(0) = sin(pi) = 0

[@feuerwelle4562]

Like i said just a mistake in the notation.

[@gustavozola7167]

Yeah.. I noticed that too

[@bprpcalculusbasics]

Ah yes. Thanks.

[@Sebastian-yi4kz]

i have a math mid-semester final exam around 8 hours from now and i watched this as a refresher since i wasn't able to practice solving IVPs as much as i would've wanted earlier today. your video helped me a lot! thank you!

[@NemoTheGlover]

thanks 💪 can't wait for series in the next 2-3 weeks

[@lesserknownfacts7849]

Please make a video discussing the domains of the following pair of functions - 1) 1/tan(x) and cot(x) 2) 1/sin(x) and cosec(x) 3) 1/sec(x) and cos(x) It would be really interesting to watch the video in which you will explain why the above pairs of functions are equal or unequal.

[@lesserknownfacts7849]

@@aashsyed1277 Ik, but sec(x) is undefined at all odd integral multiples of pi/2.

[@lesserknownfacts7849]

@@aashsyed1277 If sec(x) is undefined then 1/sec(x) also has to be undefined.

[@lesserknownfacts7849]

@@aashsyed1277 The 2 things are different. The limit exists but the function is undefined.

[@feuerwelle4562]

@@lesserknownfacts7849 you can see 1/sec(x) in two ways. 1/sec(x) and 1/1/cos(x). Multiply top and bottom by cos(x) and you get just cos(x)

[@AlgyCuber]

3:52 shouldn't it be sin(pi)?

[@bprpcalculusbasics]

Ah yes. Thankfully sin(pi)=sin(0)=0

[@sarmadmaqsood727]

loved this video even thought im in calc 1 this enlightened me nonetheless

[@ethan8559]

Good job samrad

[@sarmadmaqsood727]

@@ethan8559 🦗

[@sudhakarmathsacademy]

It is really great mathematician

[@user-wu8yq1rb9t]

Thank you

[@amanullahasif7647]

At 16:00 why don't we take e^c3 as another constant C? If we did, it would be different answer i think..

[@crosserr404]

We could do that, but the answer would be the same. y(x) = e^(c3*e^x) = (e^c3)^(e^x) = c^(e^x) So y(0) = π = c^(e^0) c = π Thus y(x) = π^(e^x)

[@hugoslavia3670]

Cheers bro

[@shannu_boi]

11:20 I'm confused. How is c3 a constant it can be +c2 or -c2? You used the positive one only and got the ans -2+2e^(1/2x^2+x-4) But if we used the negative one we get a different ans, namely -2-2e^(1/2x^2+x-4) Why do we only use the positive c?

[@feuerwelle4562]

We don't get a different result for the function. If we do it your way, we get 2 answers for C. If C is positive we get the answer C=2e^-4. If C is negative we get the solution C=-2e^-4. When we put these two into the original function we get 2e^(1/2x²+x-4)-2 and -(-e^(1/2x²+x-4))-2. As you can see, negative + negative = positive, so we actually get the same result.

[@shannu_boi]

@@feuerwelle4562 Oh right. If we use negative c then the function is is different as well. Idk why I didn't see that. Thank you very much!!

[@erikross-rnnow5517]

I'm a little confused on the first one. I only get one answer which is: -1+(1/2*sin(x)+1)^2 and not the other one. I isolated the constant before isolating y, which I guess is the reason I only have one solution and two? The two solutions bprp get also seem to be overlapping, which I recall two solutions to a DE shouldnt? Honestly prolly just me being a goof and not knowing something vital, anyone wanna help?

[@materiasacra]

You are correct, the video is wrong. There is only one (real) solution. The problem occurs at 2:56 when Lord BPRP squares both sides. This introduces a parasitic solution. If you try to substitute back the second claimed solution, you find that sqrt(y+1) is NOT equal to 1/2*sin(x)-1, which is negative for all real x, but -(1/2*sin(x)-1). Hence the second claimed solution does not solve the original equation.

[@thenew3dworldfan]

Wolfram alpha also gives 2 solutions. Which BPRP checks on wolframalpha and is correct. There is something called the existence and uniqueness theorem, that guarantees, whether or not a differential equation has multiple solutions or is guaranteed to have one solution.

[@eganrabiee627]

It's calculus...in algebra! It's Calcgebra!

[@KUDIYARASAN-]

Genius

[@Mitalee1508]

I don't why I am focusing on the pokeball he's holding 😂

[@anshumanagrawal346]

0:01 Just seperate it Jk