It look like every time he change the equation from y world to u world and then to s world, the order of the differential equation decrease by one. Could we use this trick to solve other high order differential equation problem?
@@gagadaddy8713 Wrong comment section lol. But yeah, you could. As long as your initial equation is in terms of the derivatives only, you could theoretically just keep on applying this trick until you arrive at a first order DE.
Nice job! There's actually another way you can solve this that will give you a single, explicit solution for y. If you let u = y' and v = u'/u, then the diff'l equation reduces to v' = v^2 - 1. Solving for v, then u, then y gives y = B ln |(1 + Ae^x)/(1 - Ae^x)| + C, which I believe is equivalent to your implicit solutions. No sqrts necessary!
I did it by writing the initial function in form of (y''/y')'+1=(y''/y')². Then take y''/y' = t and the solving t' by chain rule and integrating both sides. Then we can follow the method in the last part of video. Nice problem
Hey, Maths505, would you be into doing some videos involving non-elementary functions? I know you watch Dr. Penn and have seen a few there, but I'm *always* looking for more! log integral, polylog, sin/cos/exp integral, lambert W is huge, digamma is always a killer but you've done that before, etc. Anyway, great stuff as always!
this was SUCH a cool video -- but it''s KILLING ME that you didn't turn the final answers into explicit y(x) form where possible!! The second one doesn't look possible but might benefit from some weird trig identities (maybe over the complex numbers they could combine more easily?) but the first one is -b/2 + (c/2)*artanh(A*exp(x)) if i did that in my head correctly (A=e^a but its arbitrary anyway). also, kinda neat that you have one of the arbitrary constants as a literal constant term, to line up with the first trivial solution u=dy/dx=0. and how inverse trig(h) and exp "cancel" in a sense.
Great going 505. You bashed up this super villain integral like Bond 007😅. Honestly, I have never seen the hyperbolic functions popping out of integration. Enjoyed every bit of it
Great video, but it would of been nice if you went into detail how you did some of the integrals, as I feel you passed over some steps that I wasn’t sure about it. Btw I love the videos, you’re legit my favourite math problem solving channel.
Interesting, though why did you not just substitute dy/dx=1/v, which immediately gives v''=v i.e. v=Ae^x+Be^{-x} and y=int(1/(Ae^x+Be^{-x}) dx) + C=1/sqrt(AB)*arctan(sqrt(A/B)*e^x) + C.
Awesome. But I don't get why working with du/dy and ds/du instead of du/dx and ds/dx to fiddle with easier DEs and then getting the "y" only at the end.
Hi we can set y= e^integral of udx and then we have second order equation because this differential equation is homogeneous equation and then do what you did so after all , we will have first order equation.
I know that this might be wrong but I used an auxiliary function. I let y = e ^ (mx) because you can factor out the e^(2mx) that I got... I know that my repeated roots are wrong... What would be the repeated roots?
I liked your approach! I tried myself before watching and stumbled on an alternate attack that lead to explicit solutions. First, I substituted u=ln y'. The DE reduced to u'' - (u')^2 + 1 = 0. This is a first order equation in u'. I found it has solutions: u1' = -coth(x+c) u2' = -tanh(x+c) u3' = 1 u4' = -1 Integrate for possible functions for u(x): u1 = ln[A csch(x+c)] u2 = ln[A sech(x+c)] u3 = x + ln A u4 = -x + ln A When I exponentiate and integrate, I find the possible solutions for y(x) to be: y1 = A ln|tanh((x+c)/2)| + B y2 = A arctan[exp(-x-c)] + B y3 = A exp(x) + B y4 = A exp(-x) + B
Damn, this whole video felt like art rather than Math.
It look like every time he change the equation from y world to u world and then to s world, the order of the differential equation decrease by one. Could we use this trick to solve other high order differential equation problem?
@@gagadaddy8713 Wrong comment section lol.
But yeah, you could. As long as your initial equation is in terms of the derivatives only, you could theoretically just keep on applying this trick until you arrive at a first order DE.
@@daddy_myers Will try some differential equation with only y derivative and see if that really work! 😘
UND WE"RE BACC! man this was a nice refresher! maths 505 providing the freshest of diff equation
Nice job! There's actually another way you can solve this that will give you a single, explicit solution for y.
If you let u = y' and v = u'/u, then the diff'l equation reduces to v' = v^2 - 1. Solving for v, then u, then y gives
y = B ln |(1 + Ae^x)/(1 - Ae^x)| + C,
which I believe is equivalent to your implicit solutions. No sqrts necessary!
I did it by writing the initial function in form of (y''/y')'+1=(y''/y')². Then take y''/y' = t and the solving t' by chain rule and integrating both sides. Then we can follow the method in the last part of video. Nice problem
Your arrows look like the letter “S”
Hey, Maths505, would you be into doing some videos involving non-elementary functions? I know you watch Dr. Penn and have seen a few there, but I'm *always* looking for more! log integral, polylog, sin/cos/exp integral, lambert W is huge, digamma is always a killer but you've done that before, etc. Anyway, great stuff as always!
this was SUCH a cool video -- but it''s KILLING ME that you didn't turn the final answers into explicit y(x) form where possible!! The second one doesn't look possible but might benefit from some weird trig identities (maybe over the complex numbers they could combine more easily?) but the first one is -b/2 + (c/2)*artanh(A*exp(x)) if i did that in my head correctly (A=e^a but its arbitrary anyway). also, kinda neat that you have one of the arbitrary constants as a literal constant term, to line up with the first trivial solution u=dy/dx=0. and how inverse trig(h) and exp "cancel" in a sense.
wow this one is savage indeed , you had me pausing and rewinding a few
Great going 505. You bashed up this super villain integral like Bond 007😅. Honestly, I have never seen the hyperbolic functions popping out of integration. Enjoyed every bit of it
Very nice video ❤️🔥
I have a very nice Integral question
Integral from 0 to 1 [sin(xpi) * x^x * (1-x)^(1-x)]dx = ?
the answer is {pi e /24}
Pie/24, nice. 😌
Contour integration ......😌😌
Fantastic
Fantastico 😄
Great video, but it would of been nice if you went into detail how you did some of the integrals, as I feel you passed over some steps that I wasn’t sure about it.
Btw I love the videos, you’re legit my favourite math problem solving channel.
You can look up the integrals from any table of antiderivatives. They're elementary here in the video.
Interesting, though why did you not just substitute dy/dx=1/v, which immediately gives v''=v i.e. v=Ae^x+Be^{-x} and y=int(1/(Ae^x+Be^{-x}) dx) + C=1/sqrt(AB)*arctan(sqrt(A/B)*e^x) + C.
Yeah....
I also solved in a similar way
just put Y=dy/dx
Now,
Y(d²Y/dx²) + Y² = 2(dY/dx)²
By solving, we get
Y=1/(aeˣ+be⁻ˣ) = dy/dx
Awesome. But I don't get why working with du/dy and ds/du instead of du/dx and ds/dx to fiddle with easier DEs and then getting the "y" only at the end.
Hi
we can set y= e^integral of udx and then we have second order equation because this differential equation is homogeneous equation and then do what you did so after all , we will have first order equation.
and due to what I calculated y is 1 y’ is u , y’’is u’+u^2 and y’’’is u^3 +3uu’+u’’
Sounds like a nice plan
Great job. Thanks
I know that this might be wrong but I used an auxiliary function. I let y = e ^ (mx) because you can factor out the e^(2mx) that I got... I know that my repeated roots are wrong... What would be the repeated roots?
just put Y=dy/dx
Now,
Y(d²Y/dx²) + Y² = 2(dY/dx)²
By solving, we get
Y=1/(aeˣ+be⁻ˣ) = dy/dx
You should have mentioned that by introducing u =1/Y the equation for u becomes u'' - u = 0.
@@kareolaussen819
I solved it without putting Y=1/u
Coz I not got the intuition that by using this substitution it become easy
I liked your approach! I tried myself before watching and stumbled on an alternate attack that lead to explicit solutions.
First, I substituted u=ln y'. The DE reduced to
u'' - (u')^2 + 1 = 0.
This is a first order equation in u'. I found it has solutions:
u1' = -coth(x+c)
u2' = -tanh(x+c)
u3' = 1
u4' = -1
Integrate for possible functions for u(x):
u1 = ln[A csch(x+c)]
u2 = ln[A sech(x+c)]
u3 = x + ln A
u4 = -x + ln A
When I exponentiate and integrate, I find the possible solutions for y(x) to be:
y1 = A ln|tanh((x+c)/2)| + B
y2 = A arctan[exp(-x-c)] + B
y3 = A exp(x) + B
y4 = A exp(-x) + B
Great solution @sheldoc ! Thank you for taking the time to write it out, it's really helpful to see the line of thinking :)
Could you make videos like this for PDEs , thank you .
How bout solve sin(d2y/dx2)+cos(dy/dx)+y=0?
Is the zero excluded from the domain of the final function ?
Is this equation homogeneous? Second term is d2/dx2, other 2 terms are d4/dx4...
Holy click bait Batman! :D
Atleast I didn't pull a flammy😂
The DE was 🔥 though
@@maths_505 Haven't watched the video yet TBH. I'm still mulling things over. :D
And flammy boi can be a bit much some times, NGL