Solving An Interesting Differential Equation

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[@SunShine-ly7jg]

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).❤❤

[@philkazakhstan]

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

[@michaelbaum6796]

C negative leads to an inverse tanh function.

[@philkazakhstan]

@@michaelbaum6796 inverse tanh is composed of logarithms also

[@michaelbaum6796]

Thanks a lot Phil, you are right👍

[@bobbyheffley4955]

​@@philkazakhstanall the inverse hyperbolic functions involve logarithms

[@Drk950]

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!

[@danielc.martin]

So great!

[@anotherelvis]

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

[@Blaqjaqshellaq]

So y=a*tan(a*x/2+b).

[@goldfing5898]

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.

[@goldfing5898]

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).

[@vladislavlukmanov4998]

y=const is also a solution!

[@raygolu]

Yes And log y square equal to x also a solution 😂

[@user-lg6fq1yt4g]

Exactly 😅

[@bouazabachir4286]

Thanks a lot professor I follow you from Algeria.

[@SyberMath]

You are very welcome 😍

[@HoSza1]

y = 2c * tan(cx+d)

[@scottleung9587]

Very good!

[@dariosilva85]

y = 2 / (c - x) satisfies the original differential equation and it has nothing to do with tan.

[@SyberMath]

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)

[@actions-speak]

@@SyberMath It's from the zero case 🙂

[@alphastar5626]

It is either arctan (tan-¹) or argtanh (tanh-¹) depending or wether c is positive or negative

[@user-kp2rd5qv8g]

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.

[@jimschneider799]

Your solution development assumes that the first constant of integration is a positive real number, which may or may not be true.

[@SyberMath]

I agree!

[@AndresAndTheCaps]

You;re a life saver! I'm learning this in class now and it makes no sense cause my prof only uses proofs

[@SyberMath]

Thank you! Glad to hear that 😍

[@barakathaider6333]

👍

[@pyrite2060]

y=Ptan(Qx+R)

[@rob876]

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)

[@krabkrabkrab]

This is the best way to do it.

[@giuseppemalaguti435]

y=f(tgx)...y'=sec^2x...y''=2secxsecxtgx..=>y''=yy'...ovviamente bisogna sistemare i coefficenti