How to set up and solve Second-order linear ordinary differential equation

With given data:

f’’(x) = 6x , f’(2) = 6 , f(2) = -3

Thanks for your help.

f '(x) = ∫ f ' '(x) dx
= ∫ 6x dx = 3x^2 + c1
f ' (2) = 12 + c1 = 6
==> c1 = - 6
f ' (x) = 3x^2 - 6

f(x) = ∫ f ' (x) dx
= ∫ 3x^2 - 6 dx = x^3 - 6x + c2
f(2) = 8 - 12 + c2 = -3
==> c2 = 1
hence: f(x) = x^3 - 6x + 1

If you integrate the derivative you get back to the original function.

Find the general solution by direct integration:
f''(x) = 6x
f'(x) = ∫ 6x dx
f'(x) = 3x² + C₁
f(x) = ∫ (3x² + C₁) dx
f(x) = x³ + C₁x + C₂

Find the particular solution by solving for the constants:
When x = 2, f'(x) = 6
12 + C₁ = 6
C₁ = -6
When x = 2, f(x) = -3
8 + 2C₁ + C₂ = -3
C₂ = -2C₁ - 11
C₂ = 12 - 11
C₂ = 1
f(x) = x³ - 6x + 1

Take the antiderivative of f'' to get f'(x) = 3x^2 + C. Plug in the point (2,6) to find C for f'.

6 = 3*2^2 + c
6 = 12 + c
-6 = C

So f'(x) = 3x^2 - 6

Take the antideriv to get f(x) = x^3 - 6x + C.
Plug in (2, -3) to find C.
-3 = 2^3 - 6*2 + C
-3 = 8-12 + C
-3 = -4 + C
1 = C

f(x) = x^3 - 6x + 1

Jen

f[x] = -15 + 3 x^2