A Simple Yet Interesting Differential Equation: dy/dx = x+y

[@JohnSmith-mz7dh]

You can differentiate this equation twice. You get that the third derivative equals the second. This leads to y’’=me^x where m is a constant. We can integrate this once and get that y’=me^x+c. Integrate this again and we get y=me^x+cx+d If x is equal to zero, then y’ at zero is equal to y at zero. Therefore m + c= m+d which means that c = d We realise that y’-x-y=0. We therefore have remaining the equation c- x-cx-d= c-x-cx-c=x-cx= 0. c=d=-1. We therefore have y=me^x-x-1, where m is any real number.

[@Mathority1729]

This is a very neat and interesting method! Thanks so much for sharing! Really appreciate it! 😄

[@davincidamaster5311]

How do you get the initial y''=me^x equation?

[@JohnSmith-mz7dh]

@@davincidamaster5311 When we differentiate twice, we get y’’’ =y’’. The first time we differentiate we get y”=1+y’, the second time we get y’’’=y’’. We can let some function g=y’’, then g’=g. The general solution to this equation is g=y’’= me^x, where m is any real number. Ok, to get this solution, we let g=dg/dx, then multiply by dx and divide by g. We get dx=1/g dg. We integrate on both sides, and get x+c = ln|g|, solve for g, we get g= e^(x+c) = e^c*e^x=me^x = y’’.

[@potaatobaked7013]

@@davincidamaster5311 y’ = x+y take derivative of both sides y’’ = 1+y’ take derivative again y’’’ = y’’ make a substitution u = y’’ u’ = u then this is the well known identity u = me^x Therefore y’’ = me^x

[@PaulMutser]

Minor typo there at the end. Where it says...=x-cx=0... it should say ...=-x-cx=0...

[@mitcigamer4289]

how does he draw the integral symbols so perfectly

[@Mathority1729]

Hahaha your comment genuinely brightened my day this morning! Thanks for the compliment, I try 😂 And thank you so much for watching! I greatly appreciate you!

[@wdujsub7902]

Perfect crowbar

[@Mathority1729]

@@wdujsub7902 😂

[@raleu]

We need a tutorial for it

[@Bruno-yg9lu]

not just the integral sign, all his writing is majestical

[@igoranisimov6549]

This is inhomogeneous first order ODE. The solution of homogeneous equation is C*exp(x) instantly. Then we need to find a partial solution of inhomogeneous equation in the form Ax+B and add it to inhomogeneous solution.

[@Mathority1729]

Thanks for watching and for sharing the helpful comment!

[@danh5368]

This is how I solved it as well. I’d imagine this would be the most standardized way of doing it.

[@recurrenTopology]

@@danh5368 Yeah, this is the way it is taught in an undergrad ODE course.

[@frentz7]

I think you mean "first order *linear* ODE." And a couple other missteps.

[@igoranisimov6549]

@@frentz7 There are no missteps. It is obviously "linear", but it is also inhomogeneous in the form dy/dx - y = f(x), where f(x) = x. Homogeneous form is dy/dx - y = 0

[@gamemakingkirb667]

I’m taking calculus right now and you just taught me how a solution to a differential equation works; thanks!!

[@Mathority1729]

I’m so glad to hear that! Thanks a ton for watching! 😄

[@dustydrawer4494]

I forgot that u-sub works in diff equations as well lol. At least it's understandable for me.

[@Mathority1729]

Yeah, this one was a weird one, we usually don’t use u-sub for diff equations, but it works nicely here haha 😄

[@VoidFame]

Comes up again with u = 1/r in the Poisson equation

[@dustydrawer4494]

@@VoidFame wtf is that lol.

[@VoidFame]

@@dustydrawer4494 it's basically (d^2/dx^2 + d^2/dy^2 + d^2/dz^2)f(x,y,z) = g(x,y,z) Give it a look up. It's very common in physics.

[@dustydrawer4494]

@@VoidFame I did look it up. All I got was smth with the laplacian operator, which I can't understand.

[@joemcz2564]

Very nice approach! I personally like the method of subtracting both sides by y since using the product rule is a helpful tool in general. Like solving x'(t)=Ax(t)+b(t) where x and b are vectors and A is an appropriately sized square matrix. The steps to find the solution x(t) would be as follows: x'(t)= A x(t) + b(t) x'(t) + (-A) x(t)=b(t) M(t) x'(t) + (-M(t)A) x(t) = M(t)b(t) where M(t) is some matrix of same size of A setting M'(t) = -M(t)A implies M(t) = exp{-At} exp{-At} x'(t)+ d/dt[exp{-At}] x(t) = exp{-At} b(t) d/dt[exp{-At} x(t)] = exp{-At} b(t) integrate from 0 to t [exp{-At} x(t)] - [exp{-A*0} x(0)] = integral(exp{-Aτ} b(τ) dτ, from 0 to t) exp{-At} x(t) - x(0) = integral(exp{-Aτ} b(τ) dτ, from 0 to t) exp{-At} x(t) = x(0) + integral(exp{-Aτ} b(τ) dτ, from 0 to t) x(t) = exp{At}x(0) + exp{At}integral(exp{-Aτ} b(τ) dτ, from 0 to t) x(t) = exp{At}x(0) + integral(exp{A(t-τ)} b(τ) dτ, from 0 to t) When my advanced controls professor did this derivation, he didn't do the whole "let M(t) be some unknown matrix" like I did; rather he just multiplied everything by exp{-At} with a very "just trust me" attitude before using the product rule, which didn't sit right with me because I couldn't understand how someone would know to do that. Some time this year, I saw someone derive the Beltrami Identity and he also used the product rule, which finally clicked it for me. After that, this derivation of x(t) became almost obvious to me, and anytime I see some function and its derivative show up an equation in some affine sort of way, I always attempt to use the product rule.

[@Mathority1729]

Thanks for the insightful comment! That’s a fantastic method indeed, appreciate you for taking the time to detail it out for all to see and learn! And I agree, doesn’t sit right with me either whenever a teacher has a “trust me” attitude regarding a step in a problem. I want to be able to fully understand why we’re doing something! Otherwise it bothers me haha Anyways, thanks a ton for watching the video!

[@Jono98806]

You can also solve this differential equation by subtracting y on both sides and multiplying throughout by the integrating factor e^-x. Then after integrating and solving for y, you get the same solution.

[@tomrace3057]

That's how I did it. Needs integration by parts to integrate x e^(-x) but that's straightforward.

[@Psychexd]

I think this can also be done by the integrating factor method

[@Mathority1729]

Oh interesting point, I’ll have to look into and try that out! And Thanks for watching, appreciate your time!

[@jamescyriac1771]

First thing that popped in my mind when I saw the equation

[@onradioactivewaves]

If you don't use the y=x constraint and solve, you get a parabola with another one rotated 90 degrees, with the intersections being the solutions. Then just get rid of the y=x solutions for the other 2 .... Was a nice easy visualization.

[@CliffSedge-nu5fv]

Every first-order linear can be solved with an integrating factor. For this one, it includes an easy integration-by-parts, but has as many steps as the substitution trick used in this video.

[@onradioactivewaves]

@@CliffSedge-nu5fv first order O.D.E.'s yay

[@jstone1211]

DEQ's always got me. I took the class three times, I did ok in the class but I never felt like I knew how to solve them. The trick is knowing the substitution method and others as well....Friggin magic..

[@Mathority1729]

It’s amazing to see all the different ways to solve these! Thanks a ton for watching! 😄

[@CliffSedge-nu5fv]

What is the motivation for using the substitution at the beginning? 1. Why not just use the integrating factor method, which always works. 2. How did you know in advance that making that substitution would be successful?

[@frentz7]

#2 is a great question

[@Mathority1729]

1. Integrating factor method definitely works, but I wanted to share a method that anyone even in an intro calculus class would understand. Even if the method is not broadly applicable to a larger range of diff eqs, I find value in sharing something that may spark people’s interest in math. If I was purely pursuing utility, I would’ve just presented the integrating factors method, because it’s a more generally applicable solution. But that solution is already broadly taught anyways by many others. 2. I had solved it in this way many years ago before I had even come across the integrating factor method. About a decade ago, when I was still in high school, this exact differential equation showed up on a Calculus quiz, and the question didn’t require a solution, rather it wanted me select the correct graph of the slope field. But I had spare time and figured out the exact solution instead during the quiz using the technique I showed in this video. My teacher the next day was surprised like, “well I never expected anyone to actually solve this” lol. But anyways, that’s the story behind this equation for me haha Thanks for watching the video!

[@tlm3778]

this guy is so underrated

[@Mathority1729]

Haha thank you so much! 😄

[@nikolasdritsas2588]

I think that in these type of diff eq the simple way to solve them is to move y to one side and then multiplay with e^(ax) (where a will be the number that (e^(ax)*y)' will be the result of the side with y . Then we solve an easy integral from the other side so this would be y'-y=x e^(-x)*y' - e^(-x)*y = xe^(-x) (e(-x)*y)' = xe^(-x) ye^(-x) = integral of (xe^(-x)) (with respect to x ) Using DI method (blackpenredpen has an amazing video on it and it is very simple) so y = [-xe^(-x) - e^(-x) + c]/e^(-x) Now having said that it's not the only way but it is the more broad way to solve these types of diff eq. Have a good day and good luck to your channel

[@Mathority1729]

That’s a fantastic method as well, and yes indeed more broadly applicable! Thanks for watching and sharing the insightful comment for others!

[@leif1075]

What did tou mean a will be the number that eil.redult on the other side with y..infinite thst language unclear sorry..I get a is a constant but not sure what you mean..

[@nikolasdritsas2588]

​@@leif1075 sorry let me try to make it clear if you have an equation y' + a(x)*y = f(x) where f is any function of x then you can multiply by e^[integral of a(x)] (and we can ignore the constant rn) (let's call the Integral of a(x) A) so A'(x) = a(x) so it will become e^A *y' + e^A *a(x) *y = e^A * f(x) so the right hand side is [ e^(A)*y ] ' = e^A *f(x) Then we integrate both side with respect to x and we divide by e^(A) ( Note: we can't ignore the constant in this step )

[@michaelroditis1952]

for the integral of 1/(u+1) don't you need to get the absolute value? ln(|u+1|)

[@Mathority1729]

In this case, we don’t need it because we are exponentiating it to solve for u+1, which removes the log. And we don’t want to solve for |u+1|, we want the full set of u+1=e^(x+C0). The only purpose the absolute value serves is to indicate that you can’t get a real answer for the log of a negative number (though technically you can still get a complex answer). And of course, thanks so much for watching! Really appreciate it!

[@neongenesiss]

@@Mathority1729I don't really get this, can you explain?

[@neongenesiss]

How does exponentiating it remove the abs. value sign?

[@frentz7]

No the purpose of absolute values is not about whether or not you can take the logarithm of a negative number; the purpose is to get all possible solutions to the differential equation (or in more elementary cases, all possible antiderivatives). So that is, y = ln (-x) also has the derivative 1/x. If you worked more carefully you would notice that you never got c1 = 0; but then again you also divided both sides by u + 1 without worrying if this were zero, or analyzing that case. @@Mathority1729

[@frentz7]

It doesn't entirely make sense. The expression e^(c0) would be positive, and so therefore c1 would be positive, which is not what we want. We also want solutions with c1 negative, and with c1 equal to zero. @@neongenesiss

[@manaayek8091]

Im taking differential equations now and we just introduced this.

[@Mathority1729]

That’s awesome! Thanks for watching!

[@onradioactivewaves]

Good work, but it would have been really nice if you could have shown a graph at the end there.

[@Mathority1729]

You’re completely right! Didn’t even think about that. Thanks a ton for the suggestion, I will incorporate that next time. Thanks for watching and for the helpful comment!

[@onradioactivewaves]

@@Mathority1729 thank you 🙂 great content regardless. Just a quick graph can convey so much information in just a second.

[@Mathority1729]

@@onradioactivewaves yep definitely!

[@torgeirHD03]

Solved it by just adding 1 to both sides y'+1=y+x+1. Substituting u=y+x+1 we get the equation u'=u. This gives us the solution Ce^x-x-1

[@Mathority1729]

Thanks so much for watching! Very cool to see all the different ways people are solving this problem in the comments! Thanks for contributing!

[@CliffSedge-nu5fv]

What about the equation let you see that adding 1 to both sides would work?

[@torgeirHD03]

@@CliffSedge-nu5fv Most differential equations can be solved by reducing it to a known equation, which mostly comes down to having some experience with it. In this case it's relatively simple since y' and (y+x)' only differ by a constant so adding 1 to both sides we achieve (y+x+1)'=y+x+1 or u'=u.

[@nomzz1]

Very nice, I enjoyed!

[@Mathority1729]

Thank you so much for watching, I really appreciate you!

[@pcklop]

I just visualized the vector field described by the equation and noticed the vectors all line up along y=-x-1, so that must be a solution. Then I got rid of the non homogenous term x and saw that obviously ke^x is a solution to the homogenous equation, so adding that to the particular solution is a solution to the non homogeneous equation. Since the ODE is first order and the solution set y=ke^x-x-1 is one dimensional, it must be the whole solution set. Not the most rigorous way to do things but it got the job done lol.

[@Mathority1729]

That’s some very good intuition! So cool to think of it that way! And thanks a ton for watching! 😄

[@CliffSedge-nu5fv]

That is the general solution strategy called guessing. For many problems, it's all you can do. This was a first-order linear equation, so guessing isn't necessary.

[@Nutshell_Mathematica]

Great keep uploading new videos

[@Mathority1729]

Thank you so much for watching and for the support! I greatly appreciate it! I’ll definitely keep posting new videos for all of y’all 🫡

[@AllenKnutson]

Rewrite as y'-y = -x, or, (1-D)y = -x where D is differentiation. Do the bogus thing of dividing through by 1-D to get 1/(1-D) . (-x) on the RHS. Expand as (1 + D + D^2 + ...) which, applied to -x, gets you a solution -x-1+0+0+... Why is it bogus? Because 1-D isn't invertible; it has a kernel, the multiples of e^x. So the general solution will be multiples of e^x plus -x-1. What if you ignored the bogosity and tried to multiply both sides of (1-D)y = -x by 1+D+D^2+...? That infinite sum makes sense when applied to polynomials, but will blow up if you try to apply it to e^x. So as long as you say "I'm only looking for polynomial solutions" it's rigorous, and finds the unique polynomial solution -x-1.

[@greedskith]

One liner soln - add 1 on both side, and use dx+dy=d(x+y) done

[@CliffSedge-nu5fv]

Why though?

[@greedskith]

@@CliffSedge-nu5fvvery elementary manipulation in compitition -- componendo dividendo

[@wavvy94]

I only just scraped through my applied mathematics course last semester and I remember hating it, why is this making me want to tackle ODEs again 😭

[@Mathority1729]

Hahaha I’m super glad to hear the video has sparked your interest in ODEs again! 🤣 Thanks so much for watching, really appreciate it!

[@mozartjpn137]

Can you solve the problem by differentiating both sides by x and transforming it into a second-order linear differential equation?

[@Mathority1729]

I believe so! I think I saw a solution like that in the comments! Thanks again for watching! Greatly appreciate it!

[@wckewhbckea]

yo. flammable maths took ur equation

[@Mathority1729]

Wow, what a coincidence (I think not!)🤣 But oh well, looks like at least he solved it using a different method, so I can’t be too upset haha. Anyways, thanks for the support man, I really appreciate you! 🫡

[@foxlies0106]

nice I was stumped thank you.

[@Mathority1729]

@@foxlies0106 thanks for watching!

[@surendrakverma555]

Very good. Thanks 🙏

[@Mathority1729]

Thank you so much for watching! Glad you enjoyed it 😄

[@Chewy427]

rather fitting that there is an e^x term in the solution to an equation that has its own derivative

[@Mathority1729]

Beautiful isn’t it? Haha. Thanks so much for watching!

[@MaximumBan]

Can you reverse self check please?

[@23bcx]

Wow really drawing this one out

[@Mathority1729]

Thanks for watching!

[@bluethundra19]

but why when we differentiate "y" again does it give us c1e^x-1 instead of giving us x+y?

[@VivekKumar-zb2pe]

What is the value of x+y?

[@Mathority1729]

dy/dx = x+y y = c1eˣ-x-1 => c1eˣ-1 = x+y dy/dx = c1eˣ-1 = x+y Hope this helps! Thanks for watching!

[@Lucas-ub7sl]

@@Mathority1729 I'm a little confused, if we started off with the equation y = c1eˣ-x-1, then would we be able to obtain dy/dx = x+y ? Sorry if this is a dumb question, I'm still fairly new to calculus. Thanks.

[@chucksucks8640]

It is interesting that you solved that without having to come up with a solution for y' - y = x and adding it to the homogeneous solution like I was taught.

[@Mathority1729]

Yeah it seems that not many have encountered this method! Glad to see that! It’s always cool to see different solutions to the same problem. This particular solution can’t be applied as broadly, but still edifying to see nonetheless! And thanks for watching! I greatly appreciate it! 😄

[@chucksucks8640]

Have you tried doing it multiple times for things like y' = x^2 + y It seems like you have to get rid of the x variable in order to get it work. @@Mathority1729

[@DanoshTech]

Damm that’s a nice integral

[@Mathority1729]

Hahaha thank you so much 😂 Appreciate your support! Thanks for watching!

[@sazzadulkader]

Love from Bangladesh 🇧🇩❤

[@Mathority1729]

Thank you my friend! Love from the States as well! Tremendously appreciate your support :) 💙

[@garfungled7093]

y' - y = x, multiply both sides by e^(-x), reverse chain, integrate both sides, then divide by e^-x

[@Mathority1729]

Yep that works too! Thanks for watching!

[@FruitRefuter]

Integrating factor method also works

[@Mathority1729]

Yep, you're correct! Thanks for watching!

[@mtate405]

Mathority.Sounds great!

[@Mathority1729]

Thanks a ton! I spent way too long trying to come up with a good, unique name for the channel hahaha. I’m actually super glad you like it! 😄

[@Mathority1729]

Also btw fun fact, if you read out the text in my profile picture phonetically, it sounds like “I ate some Pi” 😆

[@frentz7]

your c1 = e^(c0) is flawed, as it would mean c1 > 0. (the final answer is correct; this is explained by the fact that you also failed to get the most general antiderivative of 1 / (u + 1).)

[@absolutedesi5899]

This can be done easily by the linear differential equation method

[@JunYoru-y3x]

nailed it

[@Mathority1729]

Thanks for the support! Appreciate you watching! :)

[@Dalen22_W]

why do you write it as log and not as ln?

[@Cheesy_33]

yeah, ive seen that kind of notation everywhere. not sure why though

[@Mathority1729]

Originally, I always used to write ln, but for some reason I switched to writing log at some point. I think because a lot of papers and proofs I read used log instead. I guess it’s just implied that the base is e unless otherwise explicitly stated. But yeah, ln would be the more proper thing to write But anyways, thanks for watching! Really appreciate you :)

[@redpepper74]

In this case it could be the log of any base because changing the base of a log just means multiplying by a different constant

[@Mathority1729]

@@redpepper74 yep!

[@AlphaFX-kv4ud]

I've been taught that log on its own is base 10 and that's what I've seen on every calculator I know of so I'm not sure where log on it's own being the natural log came from, or the other way around

[@isobar5857]

Why, when differentiating 'the solution' do we not get back the original equation...as a proof, so to speak? Just subscribed in the hope of an explanation?

[@mnk2961]

What writing software do you use?

[@Mathority1729]

I just use Notability! Thanks for watching! :)

[@lucasm4299]

Shouldn’t C_1 be strictly positive since an exponential is strictly positive?

[@ABHISHEKKUMAR-01024]

dy / dx = x + y => dy / dx - y = x => dy / dx + Py = Q

[@MathsPhysicshelp]

Great

[@Mathority1729]

Thank you so much for watching! 😄

[@keyan1219]

Why when you integrate u+1 is there no constant?

[@vishtoxic5928]

There is is but it's implied that that constant gets subtracted/added to the right side to make the final constant C

[@Mathority1729]

There technically is but there’s no point writing it because it’s consumed by the constant on the other side anyways

[@LipschitzHutchinson]

Why we have Laplace transforms.

[@Mathority1729]

Laplace Transform is a great method too, but figured not all would understand at the moment! Will make videos dedicated to it in the future :) Thanks for watching! Appreciate your time!

[@carultch]

Here's how it can be done with Laplace transforms. Given: y' - y = x Laplace transform, based on y(0) = u: (s - 1)*Y - u = 1/s^2 Solve for Y Y = (u*s^2 + 1)/(s^2*(s - 1)) Partial fractions: Y = A/(s - 1) + B/s^2 + C/s H-coverup for A & B: A = (u*1^2 + 1)/(1^2) = u + 1 B = (u*0^2 + 1)/(0 - 1) = -1 Reconstruct, partially clear the fraction by multiplying by s, and take the limit as s goes to infinity to find C: (u*s^2 + 1)/(s^2*(s - 1)) = (u + 1)/(s - 1) - 1/s^2 + C/s (u*s^2 + 1)/(s*(s - 1)) = (u+1)*s/(s - 1) - 1/s + C u = (u+1) - 0 + C C = -1 Partial fraction result: Y = (u + 1)/(s - 1) - 1/s^2 - 1/s Translate back to the x-domain: y = (u + 1)*e^x - x - 1 Since (u+1) is a constant, we can also put K in this position instead, and have the general solution: y = K*e^x - x - 1

[@Mathority1729]

@@carultch thanks for sharing with everyone! This will be very helpful to anyone who’s curious!

[@antaresguitar]

Amazing

[@Mathority1729]

Thank you so much for watching! 😄

[@Daniel-yc2ur]

Just use integrating factor

[@Mathority1729]

That’s a good method too! Thanks for watching!

[@alex_ramjiawan]

Why don't people use ln for log base e? log with the base omitted is log base 10.

[@NoahSebesta]

Woah. Why is the integral of dx, x + c. Wouldn't it just be c?

[@Mathority1729]

dx technically has a coefficient of 1 So the integral is ∫1dx What’s the anti-derivative of 1? It’s x. So ∫1dx = x+C And the +C term is added after integration because what’s the derivative of x+C? It’s 1, because the derivative of a constant (C) is always just 0, so that “information” was lost when deriving. So when integrating, we must assume that there was a constant. Hope that helps! Thanks for watching! :)

[@ezxd5192]

at 2:05 why not put +c on the left side?

[@Mathority1729]

It’s implied, because you’d just subtract that C from the C on the right to get the final C

[@ezxd5192]

@@Mathority1729 ah ok thanks

[@Mathority1729]

@@ezxd5192 no problem!

[@magles8348]

Linear differential equation

[@tontonbeber4555]

(s-1)y = x gen solution : y = A.exp(1.x) (gen solution is the solution of (s-1)y = 0) part sol. y = Bx+C B-Bx-C = x => B=-1, C=-1 part sol. y = -x-1 SO sol : y = -x-1 + A.exp(x) Then you need one initial value to find A For example if y(0)=0 => 0 = -1+A => A=1

[@Mathority1729]

Thanks for contributing!

[@elionsakshith3508]

Isn't this a linear differential eqn, but nice alternative method

[@Mathority1729]

Yep! It’s a method I felt most people hadn’t seen for this problem, and thought it would be interesting! Thanks so much for watching, I really appreciate you! 😄

[@CliffSedge-nu5fv]

​@@Mathority1729 Why do the substitution? I mean, what was it about the equation that let you know making that substitution would be successful? If there isn't a general class of equations where this method always works, learning this is useless.

[@lukandrate9866]

​​@@CliffSedge-nu5fvdidn't really see from where did the last implication appear but whatever

[@saptakpaul8815]

We can use integrating factor and solve it as a linear differential equation...

[@Mathority1729]

Yes that works too, thanks for watching!

[@lancer.p1]

What software do you use to make these? Great video by the way!

[@Mathority1729]

Thank you so much! Really appreciate it! I just use Notability!

[@lancer.p1]

@@Mathority1729 Oh, that's interesting. Have you tried GoodNotes? I use it for my Math lecturing

[@FenShen-us9tv]

Differential equations is much harder than calculus itself

[@Mathority1729]

Depends! But still fun regardless haha! Thanks for watching, I really appreciate it :)

[@MCLooyverse]

I'm gonna attempt this here. dy/dx = x + y d/dx dy/dx = 1 + dy/dx let z = 1 + dy/dx dz/dx = z ln(z) + C = x z = Ae^x dy/dx = Ae^x - 1 y = Ae^x - x + C Ae^x - 1 = x + Ae^x - x + C for C = -1 y = Ae^x - x - 1 for some A.

[@Mathority1729]

Thanks for posting your solution! I love seeing all the different ways people are going about this problem! Really appreciate you watching and taking the time! 😄

[@DatBoi_TheGudBIAS]

How ds every differential equation somehow end up with an e^x term

[@Mathority1729]

eˣ is sneaky like that! Your comment reminds of that Scooby Doo meme where Freddy lifts the blanket off the ghost and it reveals the usual suspect….in this case it’s eˣ 😂

[@CliffSedge-nu5fv]

It is that special function that is equal to its own derivative, so if you have a function variable and its derivative in an equation, e^x can combine with itself easily.

[@DatBoi_TheGudBIAS]

@@CliffSedge-nu5fv it always reminds of the "I'll change him" meme. She's the differential. He's e^x

[@MrShikamaruTV]

You forgot the absolute value for u+1 after integrating. Otherwise. c_1 is positive.

[@gelbkehlchen]

Solution: dy/dx = x+y |-y ⟹ y’-y = x |This is an inhomogeneous, linear differential equation of the first degree, initially the solution of the corresponding homogeneous differential equation ⟹ y’-y = 0 |+y ⟹ dy/dx = y |*dx/y ⟹ dy/y = dx |∫() ⟹ ∫dy/y = ∫dx ⟹ ln|y| = x+C |e^() ⟹ y = e^(x+C) = e^C*e^x = K*e^x |The solution of the original, inhomogeneous differential equation with variation of the constants. ⟹ y = K(x)*e^x ⟹ y’ = K’(x)*e^x+K(x)*e^x | inserted into the original, inhomogeneous differential equation ⟹ K’(x)*e^x+K(x)*e^x = x+K(x)*e^x |-K(x)*e^x ⟹ K’(x)*e^x = x |/e^x ⟹ K’(x) = x/e^x = x*e^(-x)|∫() ⟹ K(x) = ∫x*e^(-x)*dx |Solution by partial integration: ------------------------------------------ Partial integration can be derived from the product rule of differential calculus. The product rule of differential calculus states: (u*v)' = u'*v+u*v' |-u'*v ⟹ (u*v)'-u'*v = u*v' |∫() ⟹ ∫u*v'*dx = u*v-∫u'*v*dx ------------------------------------ = -x*e^(-x)+∫e^(-x)*dx = -x*e^(-x)-e^(-x)+C1 = -e^(-x)*(x +1)+C1 The whole solution is then: y = [-e^(-x)*(x+1)+C1]*e^x = -x-1+C1*e^x Sample by substituting it into the differential equation: y’ = -1+C1*e^x ⟹ Left side: -1+C1*e^x Right side: x-x-1+C1*e^x = -1+C1*e^x everything ok.

[@Mathority1729]

Thanks for posting your in-depth solution, this will be very helpful to many others! And thanks for watching the video! 😄

[@Amoeby]

y' - y = x. 1) Solve y' - y = 0 dy/y = dx y = De^x 2) Let D = D(x) Then y = D(x)e^x and y' = D'(x)e^x + D(x)e^x Go back to original equation, substitute y' and y and solve the following equation for D(x) D'(x)e^x + D(x)e^x - D(x)e^x = x D'(x) = xe^(-x) dD(x) = xe^(-x)dx D(x) = -xe^(-x) - e^(-x) + C y = (-(x + 1)e^(-x) + C)e^x y = Ce^x - x - 1

[@Mathority1729]

That works too! Thanks for watching!

[@honestadministrator]

d / dx ( x + y) = x + y + 1 ln ( x + y + 1) = x + a y = A exp ( x) - ( x + 1)

[@aureliomega3268]

dy = x + y __ dx assuming y=Ax^2+Bx+C easily can be demostrated that A=0, B=C, B+1=0, B=-1, C=-1 then: y=-x-1 done! checking: dy = -1 = x + y = x - x -1, which is correct __ dx I know is too simple... but why complicate things? 😄

[@arianlili7252]

Hi I'm from india and we are taught this type of stuff (binomial theorem, sequence and series,permutations and combinations,complex numbers,matrices, determinants ,vectors and 3d, probability,statistics, straight line,circles,hyperbola,parabola,ellipse, functions,inverse trigonometric functions,limits, continuity and differentiability,differentiation, application of derivatives, integration,area under curve, differential equation)12th grade(age 17) when is this taught in other countries?

[@ravikiran4495]

Depends, I've studied my high school(Graduated age: 16) and bachelors(Mech/aerospace)(graduated age :20) in India and as you said it was pretty math intensive,Though there was this disconnect between theoretical math and practical applications for example you learn laplace transform in the 1st year undergrad but then you dont use it right away, until you get to the second/third year of your undergrad (signal processing, heat transfer, electrical circuit analysis.control theory)etc so its not apparent until you go into more research oriented programs (Like my masters in europe) there you'll need to understand in both perspective and draw the missing link (you have to be a part time mathematician and a part time engineer and also do the job of applied math/physics) so you really need to relearn everything, anyways, that was my experience.Sorry for a longass reply and wishing you the best.

[@Daniel31216]

We also do all of that in high school in Australia.

[@Mathority1729]

I’m from America, and I was fortunate to learn all those concepts in high school as well! I’m sure it depends on the region, school, course selection, etc. too. But it’s cool to hear that it’s basically the same in India as well! And thanks for watching the video! I appreciate it :)

[@CliffSedge-nu5fv]

I promise you, no one cares.

[@что-ф3п]

y' = x+y, дифференциальное уравнение, мы пока такого не проходили

[@dekeltal]

Important to note that c1 can only be a *positive* real number.

[@Mathority1729]

C1 can be any real number actually. I assume you made your assumption because C1=e^C0. However complex values of C0 can produce negative real values of C1. And a C0 that approaches negative infinity produces a zero value for C1. And you can verify that all these solutions actually lie on the graph of the slope field. For example, let’s try the initial condition (-2,0). This would produce C1=-e^2 Which means C1=e^C0=-e^2 Therefore C0=log(-e^2) log(-e^2)=log(-1)+log(e^2) =log(-1)+2 And what is log(-1)? In other words, e to the what power gives you -1? Well that’s Euler’s identity! e^iπ=-1 So log(-1)=iπ Therefore C0=2+iπ This complex value of C0 produces a negative real value of C1=-e^2, which is a valid solution to the overall differential equation. This sort of thing actually happens many times where complex values act as an intermediary step to finding a real solution. Such is the case with the cubic formula as well! And in the case where C0 approaches negative infinity, C1 becomes zero, and the exponential term in our solution vanishes giving us a simple solution of y=-x-1, which fits as expected on the slope field graph. And of course, Thank you so much for watching and taking the time to comment!! I truly appreciate you!

[@indocesare14]

Laplace transforms go brr

[@Mathority1729]

Haha thanks for watching!

[@KSM94K]

Check this weird solution, don't even know it's correct or not e^(dy/dx)=e^x e^y [Diff each side] (e^x e^y) (d²y/dx²)=e^x e^y+e^y e^x dy/dx d²y/dx²=1+dy/dx d²y/dx²-dy/dx=1 e^-x d²y/dx²-e^-xdy/dx=e^-x d/dx(e^-x dy/dx)=e^-x e^-xdy/dx=-e^-x + C dy/dx=-1+Ce^x y=-x+Ce^x+K

[@Mathority1729]

Wow that solution is very cool, it makes perfect sense to me! Very neat, and certainly is a valid way to do it! Btw I assume the very last line in your comment showing the final result is supposed to be: y=-x+Ce^x-1 That’s the only part I think might have a typo, but all the steps you showed work out perfectly! Thanks so much for sharing your solution and for watching the video! Really appreciate you! 😄

[@KSM94K]

@@Mathority1729 thanks man, I was bored to solve it using x+y=u, so tried a weird way, never knew it works!

[@johngeverett]

How is that a 'solution'? You still don't know any value. Just a rearranged expression.

[@Mathority1729]

It wasn’t a rearranging, we had to perform an integration of the function to obtain the solution, which basically computes the area under the curve. I’ll explain further below: A differential equation explains the rate of change of a certain function or phenomenon. The solution to a differential equation results in an equation that describes, not the rates, but rather the quantities given initial conditions. For example, you might have a differential equation explaining the rate at which a pool fills up with water with respect to time. Meaning, if you plugged in the time, you would get the rate at which the water flows at that given time (maybe 10 gallons/second, at time=5 seconds), and the rate of flow is constantly changing, it might be pumping faster at the start and slows down towards the end. The solution to the differential equation will give you an equation that gives you the total quantity of water at the given the given time (perhaps 20 total gallon at time=5 seconds). This is important for many things like rocketry where you need to know how far a rocket has traveled at a given time, because as the fuel is used up, the rocket weighs less, and therefore travels further on less fuel, so it’s not a constant relationship. You have to solve a differential equation to be able to predict that. Anyways, hope that helps explain your question. Thanks for watching and for the comment!

[@danik0011]

why can't you just cancel out the d's

[@Mathority1729]

I assume you’re referring to the dy/dx? If not, please let me know. But dy/dx indicates something called differentiation, which represents a ratio of infinitesimally small changes. dy/dx basically means an infinitely small change in y, dy, over an infinitely small change in x, dx. And dy/dx is the derivative of the function x+y in this case. The d’s are not variables, rather differentiation operators. It’s the basis of differential calculus. Thanks for watching! Hopefully, I understood and answered your question!

[@skilledgaming2287]

Just cancel out the letter d

[@el-yago-pe]

LMAOOOO

[@Mathority1729]

The Differential Equations Gods would smite me if I did that 😅

[@Taokyle]

My stupid ass tells me to change the dy/dx to first principle

[@Mathority1729]

Haha 😆. Thanks for watching, hope you found it interesting!

[@wing_103]

y / x = x + y

[@Mathority1729]

If differential equations worked that way, life would be so much easier 🤣

[@eeduuxxx]

nice video bro rly enjoyed it keep going!

[@Mathority1729]

Thank you so much!! Appreciate your support and kind words!!

[@usopenplayer]

Thanks for your video! Your "e"s look like a very happy Pacman about to have a feast on those exponents. 😁🥮 I'm now going to have to review natural logs as well, apparently I'm rusty. 😳

[@Mathority1729]

Hahaha now I can’t unsee it 🤣 I’m really glad you enjoyed the video! Thanks a ton for watching!

[@vishtoxic5928]

lmaooo it kinda does.

[@AhasanAdib]

What are the apps you used to make this video?

[@Mathority1729]

Just Notability! Thanks for watching!