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Chapter 1: Problem 28

Find a function \(f\) such that \(f^{\prime \prime}(x)=x+\cos x\) and such that \(f(0)=1\) and \(f^{\prime}(0)=2\).

Short Answer

Expert verified
Question: Find the original function \(f(x)\) that satisfies the following conditions: \(f^{\prime \prime}(x)=x+\cos x\), \(f(0)=1\), and \(f^{\prime}(0)=2\). Answer: The function \(f(x)\) that satisfies the given conditions is \(f(x) = \frac{1}{6} x^3 - \cos x + 2x\).

Step by step solution

01

Integrate to find \(f^{\prime}(x)\)

Integrate \(f^{\prime \prime}(x)\) with respect to \(x\) to find \(f^{\prime}(x)\): $$f^{\prime}(x) = \int (x + \cos x) dx = \frac{1}{2} x^2 + \sin x + C_1$$ Here, \(C_1\) is the constant of integration.
02

Use \(f^{\prime}(0)=2\) to find \(C_1\)

Using the given initial condition \(f^{\prime}(0)=2\), substitute \(x=0\) in \(f^{\prime}(x)\) and solve for \(C_1\): $$2 = \frac{1}{2}(0)^2 + \sin (0) + C_1 \Rightarrow C_1 = 2$$ So, \(f^{\prime}(x) = \frac{1}{2} x^2 + \sin x + 2\)
03

Integrate to find \(f(x)\)

Integrate \(f^{\prime}(x)\) with respect to \(x\) to find \(f(x)\): $$f(x) = \int \left(\frac{1}{2} x^2 + \sin x + 2\right) dx = \frac{1}{6} x^3 - \cos x + 2x + C_2$$ Here, \(C_2\) is another constant of integration.
04

Use \(f(0)=1\) to find \(C_2\)

Using the given initial condition \(f(0)=1\), substitute \(x=0\) in \(f(x)\) and solve for \(C_2\): $$1 = \frac{1}{6}(0)^3 - \cos (0) + 2(0) + C_2 \Rightarrow C_2 = 0$$ So, \(f(x) = \frac{1}{6} x^3 - \cos x + 2x\)
05

Final solution

The function \(f(x)\) which satisfies the given conditions is: $$f(x) = \frac{1}{6} x^3 - \cos x + 2x$$

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