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Magazine Chapter 1: Problem 10
Solve the given differential equation. $$t \frac{d x}{d t}+2 x=4 e^{t}, \quad t>0$$
Short Answer
Expert verified
The general solution for the given differential equation is:
$$x(t) = 4e^t - 8\frac{e^t}{t} + 8\frac{e^t}{t^2} + \frac{C}{t^2}$$
Step by step solution
01
Identify the given differential equation's type
The differential equation provided is a first-order linear differential equation in the form:
$$t \frac{d x}{d t}+2 x=4 e^{t}$$
02
Find the Integrating Factor (IF)
The integrating factor (IF) is given by the formula:
$$IF = e^{\int P(t) dt}$$
where \(P(t)\) is the coefficient of the function, \(x\), which in this case is \(2/t\). Therefore, we have:
$$IF = e^{\int \frac{2}{t} dt}$$
Now, we compute the integral:
$$\int \frac{2}{t} dt = 2 \int \frac{1}{t} dt = 2 \ln |t| + C= \ln |t|^2$$
Thus, the integrating factor is:
$$IF = e^{\ln |t|^2} = |t|^2$$
Since \(t>0\), we can write the integrating factor as:
$$IF = t^2$$
03
Multiply the differential equation by the Integrating Factor
Now that we have calculated the integrating factor, \(IF = t^2\), we multiply the given differential equation by it:
$$t^2 (t \frac{d x}{d t} + 2x) = t^2 (4e^t)$$
Which simplifies to:
$$t^3 \frac{d x}{d t} + 2t^2 x = 4t^2 e^t$$
04
Integrate both sides
The left-hand side of the equation is now a perfect derivative of the product of the integrating factor and the function:
$$\frac{d}{d t}(t^2 x) = 4t^2 e^t$$
Now we integrate both sides of this equation with respect to \(t\):
$$\int \frac{d}{d t}(t^2 x) dt = \int 4t^2 e^t dt$$
The left integral is straightforward:
$$t^2x = \int 4t^2 e^t dt + C$$
To solve the right integral, we apply integration by parts twice.
Let the first integration by parts be:
\(u_1 = t^2\) and \(dv_1 = 4e^t dt\)
From this, we have:
\(du_1 = 2t dt\) and \(v_1 = 4e^t\)
Similarly, let the second integration by parts be:
\(u_2 = t\) and \(dv_2 = 8e^t dt\)
From this, we have:
\(du_2 = dt\) and \(v_2 = 8e^t\)
Now, apply the integration by parts formula (\(\int u dv = uv - \int v du\)) for both of these:
$$\int 4t^2 e^t dt = t^2 (4e^t) - \int 8te^t dt = 4t^2e^t - 8 \int te^t dt$$
Applying integration by parts again:
$$8 \int te^t dt = te^t(8) - \int 8e^t dt = 8te^t - 8e^t$$
Substituting back, we get:
$$\int 4t^2 e^t dt = 4t^2e^t - (8te^t - 8e^t)$$
Thus, the equation becomes:
$$t^2x = 4t^2 e^t - 8te^t + 8e^t + C$$
05
Solve for x(t)
Finally, we solve for \(x(t)\) by dividing both sides by \(t^2\):
$$x(t) = 4e^t - 8\frac{e^t}{t} + 8\frac{e^t}{t^2} + \frac{C}{t^2}$$
And this is the general solution for the given differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equation
A first-order linear differential equation is a vital concept in calculus and applied mathematics, playing a critical role in modeling a range of real-world phenomena. These equations take the general form of \( \frac{dx}{dt} + P(t) x = Q(t) \) where \( x \) is the function to be determined, \( t \) often represents time, and \( P(t) \) and \( Q(t) \) are known functions of \( t \).
In the sample exercise, \( t \frac{dx}{dt} + 2x = 4e^t \) represents a specific instance of such an equation, with \( P(t) = \frac{2}{t} \) and \( Q(t) = 4e^t \) being functions of \( t \). To solve these equations, one typically relies on methods like the Integrating Factor Method, ensuring that the solution process systematically handles the variables and functions involved. Understanding how to identify and tackle a first-order linear differential equation is fundamental for students, providing a foundation to approach more complex differential equations encountered in advanced studies.
In the sample exercise, \( t \frac{dx}{dt} + 2x = 4e^t \) represents a specific instance of such an equation, with \( P(t) = \frac{2}{t} \) and \( Q(t) = 4e^t \) being functions of \( t \). To solve these equations, one typically relies on methods like the Integrating Factor Method, ensuring that the solution process systematically handles the variables and functions involved. Understanding how to identify and tackle a first-order linear differential equation is fundamental for students, providing a foundation to approach more complex differential equations encountered in advanced studies.
Integrating Factor Method
The Integrating Factor (IF) Method is a powerful technique in differential equations, particularly useful in the realm of first-order linear differential equations. The core idea behind this method is to find an integrating factor, a function of \( t \), which, when multiplied by the original differential equation, turns the left-hand side into an exact derivative. This maneuver simplifies the process of integration.
The integrating factor, notated as IF, is usually denoted by \( e^{\int P(t)dt} \) where \( P(t) \) is the coefficient of \( x \) in the differential equation. Once the integrating factor is determined, the entire equation is multiplied by IF. After this step is completed, the left-hand side of the equation now represents the derivative of a product, which we then integrate to get the solution. This method hinges on prior knowledge of basic integration techniques and is an essential tool for solving many first-order differential equations.
The integrating factor, notated as IF, is usually denoted by \( e^{\int P(t)dt} \) where \( P(t) \) is the coefficient of \( x \) in the differential equation. Once the integrating factor is determined, the entire equation is multiplied by IF. After this step is completed, the left-hand side of the equation now represents the derivative of a product, which we then integrate to get the solution. This method hinges on prior knowledge of basic integration techniques and is an essential tool for solving many first-order differential equations.
Choosing the Correct Integrating Factor
In our textbook example, \( P(t) = \frac{2}{t} \) and we find the integrating factor to be \( t^2 \) after performing the necessary integration. It's important to correctly evaluate the integrating factor, and this requires careful integration of \( P(t) \), keeping in mind constants of integration can be omitted since they would eventually cancel out when the IF is applied.Integration by Parts
Integration by Parts is a fundamental technique in integral calculus, applicable to a wide range of problems, including those found in differential equations. It is derived from the product rule for differentiation and is used when integrating the product of two functions that are not easily integrable in their present form. The formula for Integration by Parts is \( \int u dv = uv - \int v du \), where \( u \) and \( dv \) are chosen parts of the function to be integrated.
The choice of \( u \) and \( dv \) often follows the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), which serves as a guideline to determine which function is easier to differentiate (\( u \) ) and which is easier to integrate (\( dv \) ). In our exercise, integration by parts is applied twice to integrate \( 4t^2 e^t \) effectively. This shows how the integration by parts method is applied iteratively and emphasizes the importance of strategic choices for \( u \) and \( dv \) to simplify complex integrals.
The choice of \( u \) and \( dv \) often follows the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), which serves as a guideline to determine which function is easier to differentiate (\( u \) ) and which is easier to integrate (\( dv \) ). In our exercise, integration by parts is applied twice to integrate \( 4t^2 e^t \) effectively. This shows how the integration by parts method is applied iteratively and emphasizes the importance of strategic choices for \( u \) and \( dv \) to simplify complex integrals.
