常微分方程

Solve f(x) = A ∫(0 to x) [ ∫(0 to x) (dx/[f(x)]²)] dx + Bx² + C , where A,B,C are any real constants.

If A = 0, obviously, f(x) = Bx² + C

If A ≠ 0, B,C = 0,
i.e. f(x) = A ∫(0 to x) [ ∫(0 to x) (dx/[f(x)]²)] dx

Let f(x) = axⁿ
∴axⁿ ≡ A ∫(0 to x) [ ∫(0 to x) (dx/(axⁿ)²)] dx
axⁿ ≡ A ∫(0 to x) [ ∫(0 to x) (x^(– 2n)/a²) dx] dx
axⁿ ≡ A ∫(0 to x) {[x^(– 2n + 1)/(a²(– 2n + 1))] (0 to x)} dx
axⁿ ≡ A ∫(0 to x) [x^(– 2n + 1)/(a²(– 2n + 1))] dx
axⁿ ≡ A [x^(– 2n + 2)/(a²(– 2n + 1)(– 2n + 2))] (0 to x)
axⁿ ≡ Ax^(– 2n + 2)/(a²(– 2n + 1)(– 2n + 2))
∴n = – 2n + 2
3n = 2
n = 2/3
∴a = A/(a²(– 2n + 1)(– 2n + 2))
a³ = A/((– 2n + 1)(– 2n + 2))
a = (³√A)/[³√((– 2n + 1)(– 2n + 2))]
= (³√A)/[³√((– 2(2/3) + 1)(– 2(2/3) + 2))]
= (³√A)/[³√(– 2/9)]
= – (³√(9A))/(³√2)

∴f(x) = – (³√(9A))x^(2/3)/(³√2) = – (³√(9Ax²))/(³√2)

If A,B,C ≠ 0,
f(x) = A ∫(0 to x) [ ∫(0 to x) (dx/[f(x)]²)] dx + Bx² + C
(d/dx)(f(x)) = (d/dx)[A ∫(0 to x) [ ∫(0 to x) (dx/[f(x)]²)] dx] + (d/dx)(Bx²) + (d/dx)(C)
(df(x))/dx = A ∫(0 to x) (dx/[f(x)]²) + 2Bx (∵(d/dx)[ ∫(k to x) a(x) dx] = a(x) , where k is any real constant and a(x) is any function.)
(d/dx)[(d/dx) (f(x))] = (d/dx)[A ∫(0 to x) (dx/[f(x)]²)] + (d/dx)(2Bx)
(d²f(x))/dx² = A/[f(x)]² + 2B(∵(d/dx)[ ∫(k to x) b(x) dx] = b(x) , where k is any real constant and b(x) is any function.)
(d²f(x))/dx² – A/[f(x)]² = 2B

Which is a non-linear second order differential equation.

我搵到好多個數學網站，都仲未搵到一隻款嘅second order differential equation係同上面果條second order differential equation相配合。這恐怕連數學家都仲諗到點solve。
It is likely that (d²f(x))/dx² – A/[f(x)]² = 2B has no exact elementally solution (i.e. a elementally function f(x) such that (d²f(x))/dx² – A/[f(x)]² = 2B).

Note: If a differential equation has no exact elementally solution, it does not mean that the differential equation has no solution, otherwise, it is meant that the solution of the differential equation cannot be expressed as an elementally form.

儘管如此，我非常多謝你問呢條問題。呢條問題確實令我和你都得益了不少。我無法完成整條題目，請見諒！果條differential equation我諗到點solve到時再通知你啦！

值得一提的是，If A ≠ 0, B,C = 0果部分能夠做到已算是非常好運。

The integral equation f(x) = A ∫(0 to x) [ ∫(0 to x) (dx/[f(x)]²)] dx + Bx² + C has TWO initial conditions:

f(x) = A ∫(0 to x) [ ∫(0 to x) (dx/[f(x)]²)] dx + Bx² + C
f(0) = A ∫(0 to 0) [ ∫(0 to x) (dx/[f(x)]²)] dx + B(0)² + C = C (∵∫(m to m) a(x) dx] = 0 , where m is any real constant and a(x) is any function.)
f′(x) = A ∫(0 to x) (dx/[f(x)]²) + 2Bx
f′(0) = A ∫(0 to 0) (dx/[f(x)]²) + 2B(0) = 0 (∵∫(n to n) b(x) dx] = 0 , where n is any real constant and b(x) is any function.)

解積分方程（最正路）的方法是：
1. Reduce the integral equation to a differential equation by continuous diff. the integral equation both sides.
2. Solve the differential equation.
3. Find all the initial conditions by putting special values of x in the integral equation such that the integrand becomes zero.
4. Find all the arbitrary constants by using all the initial conditions.

It is important that the solution of the integral equation should be unique (no arbitrary constants included).

good!