find the particular solution that satisfies the differential equation and the initial condition
f’’(x)=x^2 , f’(0)=4, f(0)=8
f(x)=
idk sorry
f" = x^2
f' = 1/3 x^3 + C
f'(0) = 4 ==> C=4, so
f' = 1/3 x^3 + 4
f = 1/12 x^4 + 4x + C
f(0)=8 ==> C=8, so
f(x) = 1/12 x^4 + 4x + 8
To find the particular solution that satisfies the differential equation given the initial conditions, you can follow these steps:
Step 1: Integrate the differential equation f''(x) = x^2 to find the general solution.
First, integrate the equation once:
f'(x) = ∫(x^2) dx = (1/3)x^3 + C₁,
where C₁ is the constant of integration.
Step 2: Apply the initial condition f'(0) = 4 to find the value of C₁.
Substitute x = 0 and f'(x) = 4 into the equation from Step 1:
4 = (1/3)(0)^3 + C₁
4 = C₁.
Therefore, the constant of integration C₁ is 4, and the equation becomes:
f'(x) = (1/3)x^3 + 4.
Step 3: Integrate the equation from Step 2 to find the general solution for f(x).
Integrate the equation again:
f(x) = ∫[(1/3)x^3 + 4] dx = (1/12)x^4 + 4x + C₂,
where C₂ is the new constant of integration.
Step 4: Apply the second initial condition f(0) = 8 to find the value of C₂.
Substitute x = 0 and f(x) = 8 into the equation from Step 3:
8 = (1/12)(0)^4 + 4(0) + C₂
8 = C₂.
Thus, the constant of integration C₂ is 8, and the equation becomes:
f(x) = (1/12)x^4 + 4x + 8.
Therefore, the particular solution that satisfies the differential equation f''(x) = x^2, f'(0) = 4, and f(0) = 8 is:
f(x) = (1/12)x^4 + 4x + 8.