Problem 53 - College Algebra and Calculus: An Applied Approach

Question

Find the particular solution \(y=f(x)\) that satisfies the differential equation and initial condition. $$ f^{\prime}(x)=\frac{2-x}{x^{3}}, x>0 ; \quad f(2)=\frac{3}{4} $$

Step-by-Step Solution

Answer

The particular solution which satisfies the differential equation \(f^{\prime}(x)=\frac{2-x}{x^{3}}\) and the initial condition \(f(2)=\frac{3}{4}\) is given by \[ f(x) = -\frac {1}{x^{2}} - ln|x| + 1 + ln(2) \] With \(x>0\).

Step 1: Integrate the differential

To begin with, let's find the integral of \(f^{\prime}(x)=\frac{2-x}{x^{3}}\). \n This results in \n\[ f(x) = \int f^{\prime}(x) dx = \int \frac {2-x}{x^{3}} dx \n\] This is a standard integral and it can be rewritten as \n\[ f(x) = \int 2x^{-3} dx - \int x^{-2} dx \] which further simplifies to \n\[ f(x) = -x^{-2} - ln|x| + C \] where C is the constant of integration.

Step 2: Apply the Initial Condition

Now we can apply the initial condition \(f(2) = \frac {3}{4}\). Subsituting these values into the equation gives us \(-\frac {1}{4} - ln(2) + C = \frac {3}{4}\) or \(C = \frac {3}{4} + \frac {1}{4} + ln(2) = 1 + ln(2) \)

Step 3: State the particular solution

Now substituting back the value of C into the original function, we obtain our particular solution:\[ f(x) = -\frac {1}{x^{2}} - ln|x| + 1 + ln(2) \] With domain \(x>0\)\n

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. In simple terms, it describes the rate at which something changes. The equation given in the exercise,
\( f'\(x\) = \frac{2-x}{x^3} \),
is a first-order differential equation because it involves the first derivative of the function \( f\(x\) \). Solving differential equations is crucial in many fields such as physics, engineering, and economics, as they can describe everything from how populations evolve over time to how electrical circuits work.

Initial Condition

The initial condition in a differential equation problem is a starting point that is used to find a particular solution to that equation. It often consists of values for the independent variable, here represented by \( x \), and the dependent variable, which is \( f\(x\) \) in our case. For our example, the initial condition is \( f\(2\) = \frac{3}{4} \). Initial conditions serve to anchor the general solution at a specific point, which helps in determining the constant of integration during the solving process.

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities and the area under and between curves. It is the inverse process to differentiation and is used to find functions when their rate of change is known. As seen in the exercise,
\[ f\(x\) = \int f'\(x\) \,dx = \int \frac {2-x}{x^{3}} \,dx \],
this process allows us to derive the function \( f\(x\) \) from its derivative \( f'\(x\) \). Integral calculus is essential when working with differential equations as it helps in finding the solutions.

Separation of Variables

Separation of variables is a common method for solving differential equations where the solution can be expressed as the product of two functions, each a function of one of the variables alone. To use this technique, all parts of the equation involving each variable are moved to opposite sides of the equation. However, in the provided exercise, the differential equation is already in a form that allows us to integrate directly without further separation. Still, understanding separation of variables is valuable as it can simplify more complex differential equations into a form that is easier to integrate.