Question

Solve the initial-value problem for \(x\) as a function of \(t\). \(\left(t^{2}-7 t+12\right) \frac{d x}{d t}=1,(t>4, x(5)=0)\)

Short answer

The solution is \(x = \ln\left| \frac{t-3}{2(t-4)} \right|\)."

Step-by-step solution

Re-arrange the Differential Equation

First, rearrange the given differential equation to separate the variables. The equation given is \((t^2 - 7t + 12) \frac{dx}{dt} = 1\). We want to write this as \(\frac{dx}{dt} = \frac{1}{t^2 - 7t + 12}\). Then, separate variables to get \(dx = \frac{1}{t^2 - 7t + 12} dt\).

Factor the Quadratic Expression

Factor the quadratic expression in the denominator \(t^2 - 7t + 12\) to make it easier to integrate. Observing the expression, it factors into \((t - 3)(t - 4)\). Thus, we have \(\frac{dx}{dt} = \frac{1}{(t - 3)(t - 4)}\).

Perform Partial Fraction Decomposition

Write the expression \(\frac{1}{(t - 3)(t - 4)}\) as a sum of partial fractions. Assume \(\frac{1}{(t - 3)(t - 4)} = \frac{A}{t - 3} + \frac{B}{t - 4}\). Solve for \(A\) and \(B\).

Solve for Constants in Partial Fractions

Multiply both sides by \((t-3)(t-4)\) to eliminate the denominator: \(1 = A(t-4) + B(t-3)\). Expanding and simplifying gives linear equations which you can solve: \(A + B = 0\) and \(4A + 3B = 1\). Solving these yields \(A = 1\) and \(B = -1\).

Integrate Both Sides

Now integrate both sides:\[\int dx = \int \left( \frac{1}{t-3} - \frac{1}{t-4} \right) dt\]The integrals are:\[x = \ln|t-3| - \ln|t-4| + C\]This simplifies to \(x = \ln\left|\frac{t-3}{t-4}\right| + C\).

Apply Initial Condition to Find Constant

Apply the initial condition \(x(5) = 0\). Substitute \(t = 5\) and \(x = 0\) into the integrated equation:\[0 = \ln\left| \frac{5-3}{5-4} \right| + C \Rightarrow 0 = \ln(2) + C\]Thus, \(C = -\ln(2)\).

Write Final Solution

Substitute the value of \(C\) back into the solution to get the final expression:\[x = \ln\left| \frac{t-3}{t-4} \right| - \ln(2)\]Simplify this to:\[x = \ln\left| \frac{t-3}{2(t-4)} \right|\]This is the function of \(t\) satisfying the given initial-value problem.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve the rates of change of a quantity. These equations have widespread applications in fields like physics, engineering, biology, and many other sciences. In our problem, we are given a differential equation that describes how a function of x changes with respect to the variable t.
This particular differential equation is of the form \( (t^2 - 7t + 12) \frac{dx}{dt} = 1 \). To solve it, we first need to rearrange and separate the variables, allowing us to treat each side of the equation independently. This process involves rewriting it as \( \frac{dx}{dt} = \frac{1}{t^2 - 7t + 12} \).
Understanding this concept is crucial because, in an initial-value problem, we not only determine the function \( x(t) \) but also ensure it satisfies a specific condition at a particular point. Initial-value problems often arise in cases where the system's starting state is known, and the task is to predict future behavior based on differential equations.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic method used to break down complex rational expressions into simpler fractions, which makes integration or other processes straightforward. In this exercise, once we factored the quadratic expression in the denominator as \( (t - 3)(t - 4) \), the next step was to express \( \frac{1}{(t - 3)(t - 4)} \) as the sum of two simpler fractions: \( \frac{A}{t - 3} + \frac{B}{t - 4} \).
Finding the values of A and B involves setting up a system of linear equations and solving for the constants. In our case, we multiplied through by the denominator, creating the equation: \( 1 = A(t - 4) + B(t - 3) \). Solving these equations, we found that \( A = 1 \) and \( B = -1 \), which then allows us to rewrite the original expression into something easier to integrate.
This technique is extremely useful, as it simplifies the integration process, especially for expressions that result from polynomial division or where roots are not immediately apparent.

Integration Techniques

Integration is a fundamental concept in calculus that deals with finding antiderivatives. In this problem, after using partial fraction decomposition, we needed to integrate both terms individually. The separated expression was \( \int \left( \frac{1}{t-3} - \frac{1}{t-4} \right) dt \).
Integrating each term, we get \( x = \ln|t-3| - \ln|t-4| + C \). Remembering the properties of logarithms, this simplifies to: \( x = \ln\left|\frac{t-3}{t-4}\right| + C \). This is indicative of the integration technique known as substitution, particularly useful in dealing with logarithmic forms.
After obtaining the general solution, we apply the given initial condition \( x(5) = 0 \) to calculate the constant \( C \). Solving this yields \( C = -\ln(2) \), resulting eventually in the specific solution \( x = \ln\left| \frac{t-3}{2(t-4)} \right| \), which satisfies both the differential equation and the initial condition. Understanding how to apply these integration techniques is critical for solving many problems involving differential equations.