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Chapter 4: Problem 67

Find the solution of the following initial value problems. $$f^{\prime}(x)=2 x-3 ; f(0)=4$$

Short Answer

Expert verified
Question: Given the initial value problem with the first derivative \(f^{\prime}(x) = 2x - 3\) and the initial value f(0) = 4, find the function f(x). Answer: The function that satisfies the given conditions is \(f(x) = x^2 - 3x + 4\).

Step by step solution

01

Integrate the given derivative

To find the function f(x), we need to integrate the given first derivative with respect to x: $$\int f^{\prime}(x) dx = \int (2x-3) dx.$$
02

Evaluate the integral

Now evaluate the integral on the right-hand side: $$\int (2x-3) dx = x^2 - 3x + C,$$ where C is the constant of integration.
03

Apply the initial value condition

We are given that f(0) = 4, which means when x = 0, the function has a value of 4. Substitute this into our function and solve for C: $$4 = (0)^2 - 3(0) + C.$$ From this, we can find the value of C: $$C = 4.$$
04

Write the final solution

Substitute the value of C back into our function to obtain the solution to the initial value problem: $$f(x) = x^2 - 3x + 4.$$

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