Solving An Interesting Differential Equation
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- Опубликовано: 13 фев 2024
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So great!
I tried another approach. Let z = y', then y'' = y * y' becomes z' = y * z. Chain rule: dz/dx = dz/dy * dy/dx --> z' = z * dz/dy
Our ODE turns into: z * dz/dy = y * z --> z * (dz/dy - y) = 0. That gives us 4 scenarios. The first solution is trivial: z = 0 --> y = c1 (remember: z = y'). Solving dz/dy - y = 0, we have the another 3 solutions, depending on the value of the constant c2. If c2 is 0, positive or negative, then the integral has a different solution and we got a reciprocal / tangent / logarithm function for y(x). Sorry for my english 😅 ¡Saludos desde Argentina!
Very good!
You worked only the case when C is +ve (C=a^2). If C=0 or -ve the result will be a different output function (Reciprocal function for C=0 and Logarithmic function for C is -ve).❤❤
Although kind of sketchy, allowing C to be negative (thus sqrt[C] imaginary) gives you the logarithmic solutions through the complex definitions of the logarithm. You are right about C=0, but maybe by using a limit we can avoid this
C negative leads to an inverse tanh function.
@@michaelbaum6796 inverse tanh is composed of logarithms also
Thanks a lot Phil, you are right👍
@@philkazakhstanall the inverse hyperbolic functions involve logarithms
So y=a*tan(a*x/2+b).
👍
y''=y*y' = 0.5*(y^2)'
y' = y^2+c Integrate both sides
∫ 1/(y^2+c) dy = 0.5* ∫ dx Solve by separation
1/√c *arctan(y/√c) = 0.5x+k Solve integrals
y = 2c * tan(cx+d)
Thanks a lot professor I follow you from Algeria.
You are very welcome 😍
You;re a life saver! I'm learning this in class now and it makes no sense cause my prof only uses proofs
Thank you! Glad to hear that 😍
I always found it remarkable that you can write the first derivative of the tangens function in two ways:
f(x) = tan(x) = sin(x)/cos(x)
By the the quotient rule, we obtain
f'(x)
= (cos(x)*cos(x) - sin(x)*(-sin(x)) / (cos^2(x))
= (cos^2(x) + sin^2(x)) / cos^2(x)
My school book then applied the addition theorem in the numerator and gave
= 1/cos^2(x)
but completely ignored the other version, to split up the fraction using the distribution law:
= cos^2(x)/cos^2(x) + sin^2(x)/cos^2(x)
= 1 + (sin(x)/cos(x))^2
= 1 + tan^2(x)
which is more intersting because it solves the differential equation
y' = 1 + y^2
If you derive that, you get
y'' = 2*y*y'
Which is very similar to the ODE in this video.
To account for the factor 2, I would generalize the tangens function to
y(x) = a * tan(bx + c)
And bulid its derivatives:
y'(x)
= a * (1 + b*tan^2(bx + c))
= a + a*b*tan^2(bx + c)
y''(x) = a*b* 2*tan(b*x + c) * (1 + b*tan^2(bx + c))
And compare that to
y(x) * y'(x) = a*tan(bx + c) * a*(1 + b*tan^2(bx + c))
Hmmm, 2*b = 1, thus b = 1/2 or a = 0 fulfills the equation.
I think it is a = 0 (trivial solution, the zero function) or a = 1 and b = 1/2, thus
y(x) = 0
or
y(x) = tan(x/2 + c).
y=const is also a solution!
Yes
And log y square equal to x also a solution 😂
Exactly 😅
It is either arctan (tan-¹) or argtanh (tanh-¹) depending or wether c is positive or negative
y=Ptan(Qx+R)
d^2y/dx^2 - v dv/dy where v = dy/dx. Thus vdv/dy = vy > v = 1/2 y^2 + c = 1/2(y^2 + a^2) > dy/(y^2 + a^2) = 1/2 dx > 1/a arctan(y/a) = 1/2 x + K. Thus, y = a tan(ax/2 + b), where a and b are constants.
y = 2 / (c - x) satisfies the original differential equation and it has nothing to do with tan.
that must be one of the values that come from fact that the integration constant can be negative (I assumed it's positive and solved accordingly)
@@SyberMath It's from the zero case 🙂
Your solution development assumes that the first constant of integration is a positive real number, which may or may not be true.
I agree!
let v = y'
then vdv/dy = yv
so v = o => y = const
or dv/dy = y => v = y^2/2 + const
=> dy/dx = y^2/2 + const
=> 2dy/(y^2 + c) = dx
=> 2∫dy/(y^2 + c) = x + a
=> 2arctan(y/b) = bx + d
=> arctan(y/b) = bx/2 + e
=> y/b = tan(bx/2 + e)
=> y = b tan(bx/2 + e)
check: y' = 1/2 b^2 sec^2(bx/2 + e)
y'' = 1/2 b^3 tan(bx/2 + e) sec^2(bx/2 + e)
yy' = 1/2 b^3 tan(bx/2 + e) sec^2(bx/2 + e)
This is the best way to do it.
y=f(tgx)...y'=sec^2x...y''=2secxsecxtgx..=>y''=yy'...ovviamente bisogna sistemare i coefficenti