Problem 32 - Advanced Engineering Mathematics

Question

$$ \begin{aligned} &\left(1+t^{2}\right) \frac{d x}{d t}+x=\tan ^{-1} t, \quad x(0)=4\\\ &\text { [Hint: In your solution let } \left.u=\tan ^{-1} t .\right] \end{aligned} $$

Step-by-Step Solution

Answer

The solution is: \(x(t) = e^{-\tan^{-1} t} \left(\int \frac{\tan^{-1} t}{1+t^2} e^{\tan^{-1} t} dt + 4 \right)\).

Step 1: Identify the Type of Equation

The given differential equation \((1+t^2) \frac{dx}{dt} + x = \tan^{-1} t\) is a first-order linear differential equation. We'll solve it by first identifying the integrating factor method.

Step 2: Rewrite the Equation in Standard Form

The standard form of a first-order linear differential equation is \(\frac{dx}{dt} + P(t)x = Q(t)\). Divide the entire equation by \(1+t^2\) to get \(\frac{dx}{dt} + \frac{1}{1+t^2}x = \frac{\tan^{-1} t}{1+t^2}\).

Step 3: Identify Integrating Factor

The integrating factor \(\mu(t)\) is given by \(e^{\int P(t) dt}\), where \(P(t) = \frac{1}{1+t^2}\). Thus, \(\mu(t) = e^{\int \frac{1}{1+t^2} dt} = e^{\tan^{-1} t}\).

Step 4: Multiply by the Integrating Factor

Multiply the entire differential equation by the integrating factor \(e^{\tan^{-1} t}\). So the equation becomes \(e^{\tan^{-1} t} \frac{dx}{dt} + \frac{1}{1+t^2} e^{\tan^{-1} t} x = \frac{\tan^{-1} t}{1+t^2} e^{\tan^{-1} t}\).

Step 5: Recognize as Exact Differential

The left side of the equation can now be written as the derivative of a product: \(\frac{d}{dt}(e^{\tan^{-1} t} x) = \frac{\tan^{-1} t}{1+t^2} e^{\tan^{-1} t}\).

Step 6: Integrate Both Sides

Integrate both sides with respect to \(t\): \(\int \frac{d}{dt}(e^{\tan^{-1} t} x) dt = \int \frac{\tan^{-1} t}{1+t^2} e^{\tan^{-1} t} dt\). Thus, \(e^{\tan^{-1} t} x = \int \frac{\tan^{-1} t}{1+t^2} e^{\tan^{-1} t} dt + C \).

Step 7: Apply Initial Condition

Use the initial condition \(x(0) = 4\) to find \(C\). When \(t = 0\), \(\tan^{-1}(0) = 0\), so the equation becomes \(e^0(4) = \int_0^0{\frac{0}{1+t^2}e^0 dt} + C\), giving \(C = 4\).

Step 8: Solve for x(t)

Isolate \(x\): \(x(t) = e^{-\tan^{-1} t} \left(\int \frac{\tan^{-1} t}{1+t^2} e^{\tan^{-1} t} dt + 4 \right)\).

Step 9: Final Solution

The solution is \(x(t) = e^{-\tan^{-1} t} \left(\int \frac{\tan^{-1} t}{1+t^2} e^{\tan^{-1} t} dt + 4 \right)\). This expression represents the solution implicitly in terms of \(t\), as the integral cannot be further simplified without specific techniques or numerical methods.

Key Concepts

Integrating Factor

An integrating factor is a powerful tool used to solve first-order linear differential equations. It's particularly useful for equations of the form \( \frac{dx}{dt} + P(t)x = Q(t) \). The integrating factor, usually denoted as \( \mu(t) \), is designed to simplify the differential equation into an exact differential form. This allows us to easily integrate and solve the equation. To find the integrating factor, we compute:

• \( \mu(t) = e^{\int P(t) dt} \)

In our specific problem, after dividing the equation by \( 1+t^2 \), we identify \( P(t) \) as \( \frac{1}{1+t^2} \). This results in the integrating factor \( \mu(t) = e^{\tan^{-1} t} \). By multiplying the entire equation by this integrating factor, we convert it into a form that can be straightforwardly integrated, leading us toward the solution. The critical aspect of using an integrating factor lies in recognizing these compatible forms and applying it systematically.

Initial Condition

Initial conditions are specifications provided along with differential equations that allow us to find particular solutions rather than just general ones. They are essential in situations where there are infinitely many solutions, as they help pinpoint the exact solution that applies to a given scenario.

• The condition is typically presented as \( x(t_0) = x_0 \), where \( t_0 \) is a specified initial time and \( x_0 \) is the function value at that time.

In our example, we are given the initial condition \( x(0) = 4 \). This initial condition allows us to calculate the constant of integration \( C \) after integrating both sides of the differential equation. We substitute \( t = 0 \) into the partially solved equation and solve for \( C \). It's a necessary step to determine a unique solution for the problem, rather than a family of solutions.

Exact Differential

An exact differential is a specific type of differential equation where the equation can be expressed as the exact derivative of a function. This transformation allows for straightforward integration to solve the equation. It follows from the structure of an exact equation that the result of integration represents the solution to the original differential equation. In our context, after determining and applying the integrating factor \( e^{\tan^{-1} t} \), the form of the differential equation on the left-hand side becomes \( \frac{d}{dt}(e^{\tan^{-1} t} x) \). This shows that it is indeed an exact differential. Hence, this property allows us to integrate directly without additional manipulation. Recognizing an exact differential can significantly simplify solving process, making it easier to find the function \( x(t) \).

Standard Form

The standard form of a first-order linear differential equation is necessary for utilizing methods like integrating factors effectively. The general structure is \( \frac{dx}{dt} + P(t)x = Q(t) \), where \( P(t) \) and \( Q(t) \) are known functions of \( t \). Placing the equation into this form is often a critical first step when solving with advanced techniques.

• Reformulating the equation in standard form helps to identify \( P(t) \) clearly, which is crucial for calculating the integrating factor.

In the provided exercise, the transformation involves dividing the entire original equation by the coefficient \( 1+t^2 \) of \( \frac{dx}{dt} \), simplifying it to \( \frac{dx}{dt} + \frac{1}{1+t^2}x = \frac{\tan^{-1} t}{1+t^2} \). Doing so sets the stage for subsequent steps in the solution process and ensures that the differential equation can be tackled effectively using standard methods.