Can You Solve An Interesting Differential Equation?
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- Опубликовано: 6 ноя 2024
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Great❤
Nice job - I knew I wasn't gonna solve it if you were gonna talk about it for 13 min, but it was fun watching you walk us thru the steps nonetheless.
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Hmm, reminds me a touch of Halley's method for approximating roots.
There is a standard way to solve this equation. Notice that the independent variable is not explicitly in the equation. Let y'=p
y''=(dp/dx)(dy/dy)
=(dp/dy)(dy/dx)
y"=pdp/dy
This will reduce the equation to a first order differential equation without resorting to tricks.
So what would y be? Integral of p + c? I dont see how that is easier
Substitute back, so that it will eventually give us the same equation *y' = 2√y + k.* And yeah, that is a standard way to solve such differential equations (without the independent variable explicitly present).
👍
Why aren’t you using differential / leibniz notation. Using lagranges notation is pretty confusing.
6:08 4=y?
I think you misunderstood the notation, y-prime is different to y^1, although they look very similar
Can I have some help? I got y=(x+c1/2)^2
Never mind I figured it out- I was doing it with respect to y
y''/y' = 1/√y
∫dy' = ∫dy/√y
y' = 2√y + c
∫dy/(2√y + c) = ∫dx
let u = 2√y + c
du = dy/√y
1/2∫(1 - c/u)du = x + a
1/2(u - c ln u) = x + a
1/2(2√y + c - c ln(2√y + c)) = x + a
(2√y + c)/2 - c/2 ln (2√y + c) = x + a