A spicy 2nd order non-linear differential equation

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This was a fun non-linear differential equation with solution development featuring an equation convertible into an exact differential form.

[quentinrenon]

To be clear, at 5:28 it's missing a factor of 1/2, also at the very end, isn't the sqrt(C) also a constant? And sinh being an odd function, you could get the +- out, and again put it with the constant, then rewrite the entire thing as:
y = e^(c*sinh(2x+c_1))

[truIyepic]

This was pretty awesome. Also, I have a suggestion: I think it'd be cool to see specific differential equations that arise in physics and maths in videos, like the Schrodinger equation for certain QM systems or Maxwell's equations. Thanks again for making these videos, they really make my day :)

[thomasstokes]

Alternatively you can set y = e^u where u is a function of x. This gives y' = u'e^u and y'' = (u''+u'^2)e^u.
Substituting into the original equations gives (u''+u'^2)e^{2u} - u'^2e^{2u} = 4ue^{2u}.
A lot of stuff cancels and you are left with u'' = 4u.
Solving: u(x) = Ae^{2x}+Be^{-2x} => y = e^{Ae^{2x}+Be^{-2x}}.

[djridoo]

I don't understand why the antiderivative of -4ln(y)/y wasn't -2ln²(y), but it was a very cool video !

[Teja]

My attempt: (y'/y)' = y''/y - (y'/y)^2 - which is LHS/y^2. But y'/y = (ln y)'. Hence, the equation simplifies to (ln y)'' = 4 ln y.
Hence, ln y = A exp(2x) + B exp(-2x).

[s.rehman.]

There is a nice and clean solution to the problem. Substitute lny = u.
it transforms it into (D^2 - 4)u = 0.
And that is u = c1e^(2x) + c2e^(-2x)
which is
lny = c1e^(2x) + c2e^(-2x)

[Risuchan]

Alternative solution: Notice that the lhs of the original eq is the derivative of a quotient (y y" - y'²) = y² (y' / y)'
Hence the equation nicely simplifies to (y' / y)' = 4 lny
Then y' / y is the derivative of lny
(ln y)" = 4 lny
From there, after substitution w = lny, we find immeditaly find w = a exp(2x) + b exp(-2x)
Finally y = exp (a exp(2x) + b exp(-2x))

[banjo]

Nice video; question, is the integral (-4*integral of ln(y)/y) at 5:28 correct? I feel like it misses a factor of 1/2 so it should be -2 ln²(y), or am I mistaken?

[ungpbll]

yy''-y'^2=4y^2*logy
y''/y-(y'/y)^2=4logy

Notice that the left-hand side is the second derivative of logy

(logy)''=4logy

logy=Ae^(2x)+Be^(-2x)
y=exp(Ae^(2x)+Be^(-2x)),

which could also be written

y=A^e^(2x)*B^e^(-2x)

[kristim]

This guy is five'0'fantastic.
Really love your content, keep going!

[jeffburrell]

What am I missing? My math chops are not what they once were, but at 9:10 should g(y) not be -2*ln^2(y)?

[MrDerpinati]

differential equations are fun to watch

[bart]

Surely you can absorb the square root of C into the constant?

[Circuito]

This was very cool, can you maybe show and solve some advanced physics/mathematics important PDE or non linear ODE like for example the Schrödinger equation for H2 particle, ora Maxwell's equations or Laplace equation etc?

[ahsgdf]

@ahsgdf13 hours ago

Why not simply let y=exp(z) leading to the ODE z'' = 4 z with the solution z = A exp(2x)+B exp(- 2x)?

[Mario_Altare]

Hi! Here's my attempt of solution:

I've divided both members by y^2:

[y''y - (y')²]/y² = 4 ln y

It follows that:

d (y'/y)/dx = 4 ln y

So
d² (y'/y) = 4 y'/y

Let y'/y = u

So u'' = 4u

u = c1e^(2x)+c2e^(-2x)

y'/y = c1e^(2x)+c2e^(-2x)

Integrating both members, we'll get

ln y = c3 + c1e^(2x)/2 - c2e^(-2x)/2

y= e^[c3 + c1e^(2x)/2 - c2e^(-2x)/2]

Not sure it works, anyway....

[zunaidparker]

7:35 aren't you assuming here that c is positive? Because if you allow for c to be negative you should get a standard trigonometric function as the anti-derivative.
Unless I'm missing some other restriction on c, I think you should carry forward both cases in order to ensure you have the complete solution no?

[jonathan]

Does this differential equation arise from problems in physics or mathematics?

[Homayoun]

Hi. Thanks for the interesting video. I've got a question regarding this equation. What if you divide both sides by y^2 and subsequently represent the LHS as d/dx(y'/y)? On the RHS remains only the term 4lny and the equation can be solved rather easily.