f"(x)=2, f'(2)=5, f(2)=10.
solve this differential equation by finding the particular solution.
How do you do this?
To solve a differential equation, we need to integrate it. In this case, you have been given a second-order differential equation and specific initial conditions. Let's start by integrating the given derivative f"(x) = 2.
We integrate f"(x) with respect to x:
∫ f"(x) dx = ∫ 2 dx
Integrating the right side, we get:
f'(x) = 2x + C₁
Here, C₁ is the constant of integration.
Now, we're given f'(2) = 5, which means when x = 2, f'(x) = 5. Let's substitute this information into the equation:
5 = 2(2) + C₁
5 = 4 + C₁
C₁ = 5 - 4
C₁ = 1
We also know f(2) = 10. Let's find the particular solution by integrating again:
∫ f'(x) dx = ∫ (2x + 1) dx
Integrating the right side, we get:
f(x) = x² + x + C₂
Here, C₂ is the constant of integration. Now, we can substitute the point (x, f(x)) = (2, 10) into the equation:
10 = (2)² + 2 + C₂
10 = 4 + 2 + C₂
C₂ = 10 - 4 - 2
C₂ = 4
Finally, we can write the particular solution:
f(x) = x² + x + 4
Therefore, the particular solution to the given differential equation is f(x) = x² + x + 4.