Problem 13 - Advanced Engineering Mathematics

Question

In Problems 9-16, solve the given differential equation. $$ t \frac{d Q}{d t}+Q=t^{4} \ln t $$

Step-by-Step Solution

Answer

The solution is \( Q = \frac{1}{5}t^4 \ln t - \frac{1}{25}t^4 + \frac{C}{t} \).

Step 1: Identify the Type of Differential Equation

This differential equation is linear with respect to the dependent variable, which is \( Q \). This equation can be expressed in the standard form \( \frac{dQ}{dt} + \frac{1}{t}Q = t^3 \ln t \).

Step 2: Find the Integrating Factor

To solve the linear differential equation, we first need to find the integrating factor \( \mu(t) \). It is given by the expression \( e^{\int P(t) \, dt} \), where \( P(t) = \frac{1}{t} \). Therefore, the integrating factor is \( \mu(t) = e^{\int \frac{1}{t} \, dt} = e^{\ln t} = t \).

Step 3: Multiply through by the Integrating Factor

Multiply every term in the differential equation by the integrating factor \( t \) to make the left-hand side an exact derivative: \( t \left( \frac{dQ}{dt} \right) + \frac{1}{t}Q \cdot t = t^3 \ln t \cdot t \), simplifying to \( \frac{d}{dt}(tQ) = t^4 \ln t \).

Step 4: Integrate Both Sides

Integrate both sides of the equation \( \frac{d}{dt}(tQ) = t^4 \ln t \) with respect to \( t \). The left side integrates to \( tQ \), and for the right side, the integral of \( t^4 \ln t \) can be found using integration by parts.

Step 5: Use Integration by Parts

Let \( u = \ln t \) and \( dv = t^4 \, dt \). Then \( du = \frac{1}{t} \, dt \) and \( v = \frac{t^5}{5} \). Using integration by parts: \( \int t^4 \ln t \, dt = \frac{t^5}{5} \ln t - \int \frac{t^5}{5} \cdot \frac{1}{t} \, dt = \frac{t^5}{5} \ln t - \frac{1}{5} \int t^4 \, dt \).

Step 6: Simplify the Right-Hand Side

Complete the calculation of the integral: \( \frac{t^5}{5} \ln t - \frac{1}{5} \frac{t^5}{5} = \frac{t^5}{5} \ln t - \frac{t^5}{25} \).

Step 7: Solve for Q

Now, we have \( tQ = \frac{1}{5}t^5 \ln t - \frac{1}{25}t^5 + C \), where \( C \) is the constant of integration. Therefore, \( Q = \frac{1}{5}t^4 \ln t - \frac{1}{25}t^4 + \frac{C}{t} \).

Key Concepts

Integrating Factor

To solve linear differential equations like the one given in the exercise, we often use a handy tool called an "integrating factor."
This technique helps us make the equation easier to work with.
The idea is straightforward: we want to multiply the entire differential equation by a certain function (the integrating factor) to transform it into a form where the left-hand side becomes an exact derivative.

The integrating factor is calculated using the form:

• For an equation in the form \( \frac{dQ}{dt} + P(t)Q = g(t) ,\), the integrating factor \( \mu(t) \) is \( e^{\int P(t) \, dt} \).
• In our example, \( P(t) \) is \( \frac{1}{t} \), which leads to \( \mu(t) = e^{\ln t} = t \).

This transformation simplifies the equation, allowing us to rewrite it as an exact derivative on the left side, making integration a feasible next step.
By multiplying the whole equation by \( t \), the left becomes \( \frac{d}{dt}(tQ) \), which is the exact derivative we need. It systematically sets us up for the integration step.

Integration by Parts

When we reached the integration step in the exercise solution, we had the term \( \int t^4 \ln t \, dt \) on the right side.
This particular integral requires a special technique known as "integration by parts."
Here's how it works in simple terms:

Integration by parts is a method derived from the product rule of differentiation.
It helps us integrate products of two different types of functions.
The formula is:

• \( \int u \, dv = uv - \int v \, du \)

In this scenario:

• Let \( u = \ln t \) and \( dv = t^4 \, dt \).
• This gives us \( du = \frac{1}{t} \, dt \) and \( v = \frac{t^5}{5} \).
• Plug these into the integration by parts formula, and simplify to get the required integral form.

This process helps us solve complex integrations by breaking them down into simpler parts, making it accessible to obtain the desired solution.

Exact Derivative

An exact derivative, in simple terms, means we have rewritten a differential equation in such a way that it resembles the derivative of a product of functions.
For our exercise, after finding the integrating factor and using it effectively, we aim to reshape the equation to exhibit this property.

Specifically, in our problem, multiplying by the integrating factor transformed the equation into:

• \( \, t \frac{dQ}{dt} + Q = t^3 \ln t \cdot t \)
• which simplifies the left side to \( \frac{d}{dt}(tQ) \)

Here, \( \frac{d}{dt}(tQ) \) represents an exact derivative.
This rewriting is useful because once we recognize the left side as an exact derivative, integrating both sides directly aids in solving for the desired function.
Consequently, we conclude by simply integrating to revert the equation to a practically usable form to extract the solution.