Find the particular solution that satisfies the differential equation and the initial condition.
f '(s) = 10s − 4s3, f(3) = 7
&
f ''(x) = 6, f '(2) = 14, f(2) = 19
To find the particular solution that satisfies the given differential equation and initial condition, we need to integrate the equation and solve for any unknown constants using the initial condition.
Let's solve the first problem:
1. Start by integrating both sides of the given differential equation:
∫f'(s) ds = ∫(10s - 4s^3) ds
2. Integrating f'(s) gives us:
f(s) = 5s^2 - s^4 + C1
where C1 is the constant of integration.
3. Now, substitute the initial condition f(3) = 7 into the equation:
7 = 5(3)^2 - (3)^4 + C1
7 = 45 - 81 + C1
7 = -36 + C1
C1 = 43
4. Substitute the value of C1 back into the equation:
f(s) = 5s^2 - s^4 + 43
So, the particular solution that satisfies the differential equation f'(s) = 10s - 4s^3 and the initial condition f(3) = 7 is f(s) = 5s^2 - s^4 + 43.
Now let's solve the second problem:
1. Start by integrating both sides of the given differential equation:
∫f''(x) dx = ∫6 dx
2. Integrating f''(x) gives us:
f'(x) = 6x + C2
where C2 is the constant of integration.
3. Solve for the constant C2 using the initial condition f'(2) = 14:
14 = 6(2) + C2
14 = 12 + C2
C2 = 2
4. Substitute the value of C2 back into the equation:
f'(x) = 6x + 2
5. Integrate f'(x) to find f(x):
∫f'(x) dx = ∫(6x + 2) dx
f(x) = 3x^2 + 2x + C3
where C3 is the constant of integration.
6. Solve for the constant C3 using the initial condition f(2) = 19:
19 = 3(2)^2 + 2(2) + C3
19 = 12 + 4 + C3
C3 = 3
7. Substitute the value of C3 back into the equation:
f(x) = 3x^2 + 2x + 3
So, the particular solution that satisfies the differential equation f''(x) = 6, f'(2) = 14, and f(2) = 19 is f(x) = 3x^2 + 2x + 3.
To find the particular solution that satisfies a given differential equation and initial condition, you can follow these steps:
Step 1: Integrate the differential equation to find the general solution.
Step 2: Use the initial condition to find the particular solution.
Let's solve each problem step by step:
1. f '(s) = 10s − 4s³, f(3) = 7:
Step 1: Integrate the differential equation with respect to s:
∫ f '(s) ds = ∫ (10s − 4s³) ds
This gives us the general solution:
f(s) = 5s² - s⁴ + C
Step 2: Use the initial condition f(3) = 7 to find the particular solution:
f(3) = 5(3)² - (3)⁴ + C = 7
Simplifying the equation, we get:
45 - 81 + C = 7
-36 + C = 7
C = 7 + 36
C = 43
Therefore, the particular solution that satisfies the given differential equation and initial condition is:
f(s) = 5s² - s⁴ + 43.
2. f ''(x) = 6, f '(2) = 14, f(2) = 19:
Step 1: Integrate the differential equation two times to find the general solution.
First integration yields: f '(x) = ∫ 6 dx = 6x + C1
Second integration yields: f(x) = ∫ (6x + C1) dx = 3x² + C1x + C2
This gives us the general solution:
f(x) = 3x² + C1x + C2
Step 2: Use the initial conditions f '(2) = 14 and f(2) = 19 to find the particular solution:
f '(x) = 6x + C1
f '(2) = 6(2) + C1 = 12 + C1 = 14
Simplifying the equation, we get:
C1 = 14 - 12
C1 = 2
Substituting the value of C1 in the general solution:
f(x) = 3x² + 2x + C2
Next, using the initial condition f(2) = 19:
f(2) = 3(2)² + 2(2) + C2 = 12 + 4 + C2 = 16 + C2 = 19
Simplifying the equation, we get:
C2 = 19 - 16
C2 = 3
Therefore, the particular solution that satisfies the given differential equation and initial conditions is:
f(x) = 3x² + 2x + 3.
f '(s) = 10s − 4s^3
f(s) = 5s^2 - s^4 + C
f(3)=7, so
5*9-81+C = 7
Do the same for the 2nd one, but use both specified conditions to find the constants.