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Find the general solution of the first-order, linear equation. $$ t x^{\prime}=4 x+t^{4} $$

Short Answer

Expert verified
The general solution is \( x(t) = |t|^{4} (\ln|t| + C) \).

Step by step solution

01

Write the Equation in Standard Form

Given the equation: \( t x^{\prime} = 4x + t^4 \). First, we divide the entire equation by \( t \) to get it into the standard form of a linear differential equation. The standard form is \( x' + P(t)x = Q(t) \). After dividing by \( t \), the equation becomes \( x' - \frac{4}{t}x = t^3 \).
02

Find the Integrating Factor

The integrating factor \( \mu(t) \) is calculated using the formula \( \mu(t) = e^{\int P(t) \, dt} \), where \( P(t) = -\frac{4}{t} \). Thus, \( \mu(t) = e^{\int -\frac{4}{t} \, dt} = e^{-4\ln|t|} = |t|^{-4} \).
03

Multiply Through by Integrating Factor

Multiply through the entire differential equation by the integrating factor \( |t|^{-4} \): \( |t|^{-4} x' - |t|^{-4} \frac{4}{t} x = |t|^{-4} t^3 \). This simplifies the left-hand side to the derivative of the product \( \frac{d}{dt}(|t|^{-4} x) \).
04

Integrate Both Sides

Now integrate both sides with respect to \( t \). We have: \( \int \frac{d}{dt}(|t|^{-4} x) \, dt = \int |t|^{-1} \, dt \). The left side becomes \( |t|^{-4} x \) and the right side integrates to \( \ln|t| + C \), where \( C \) is the integration constant.
05

Solve for the General Solution

Solve for \( x \) in terms of \( t \) from \( |t|^{-4} x = \ln|t| + C \): \( x(t) = |t|^{4} (\ln|t| + C) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard form of linear differential equation
A first-order linear differential equation is typically expressed in the standard form which is relatively easy to work with. The standard form of such an equation is written as \[ x' + P(t)x = Q(t) \]where:
  • \(x'\) represents the derivative of \(x\) with respect to \(t\)
  • \(P(t)\) is a function of \(t\) that multiplies \(x\)
  • \(Q(t)\) is another function of \(t\)
To convert a given equation into this form, we may need to rearrange or divide all terms by a coefficient. For instance, for our equation \( t x^{\prime} = 4x + t^4 \), dividing each term by \( t \) rearranges the equation into \( x' - \frac{4}{t}x = t^3 \). This puts the equation in standard form, making it ready for the next steps in the solution process.
Integrating factor method
The integrating factor method is a powerful technique to solve first-order linear differential equations. This method involves calculating an integrating factor, \( \mu(t) \), that simplifies the equation into a form that can be easily integrated. The integrating factor is calculated using the expression:\[\mu(t) = e^{\int P(t) \, dt}\]where \( P(t) \) is the function of \( t \) from the standard form. For the equation \( x' - \frac{4}{t}x = t^3 \), \( P(t) = -\frac{4}{t} \). Integrating this gives \( \int -\frac{4}{t} \, dt = -4\ln|t| \). Thus, the integrating factor becomes \( \mu(t) = e^{-4\ln|t|} = |t|^{-4} \). Once this factor is determined, the entire equation is multiplied by it, which consolidates terms, allowing us to recognize and therefore integrate the resulting expressions more easily.
General solution
The solution to a differential equation that contains an arbitrary constant is known as the general solution. After using the integrating factor to simplify the differential equation, the next step involves integrating both sides to find a solution that includes an unknown constant, often denoted \(C\).For the equation \( x' - \frac{4}{t}x = t^3 \), after multiplication by the integrating factor \( |t|^{-4} \), the equation becomes:\[ \frac{d}{dt}(|t|^{-4} x) = |t|^{-1} \]Integrating both sides gives:\[ |t|^{-4} x = \int |t|^{-1} \, dt = \ln|t| + C \]Rearranging this to solve for \(x\) yields the general solution:\[ x(t) = |t|^{4} (\ln|t| + C) \]This solution represents a family of functions, each curve differing by the value of \(C\). The presence of \(C\) accounts for any initial conditions that could tune the solution to a specific instance.
Integration techniques
Integration techniques play a crucial role in solving differential equations, especially when computing the effect of the integrating factor on the differential equation form. In our scenario, the integral \( \int |t|^{-1} \, dt \) arises after simplifying the equation using the integrating factor. This integral is a standard form, yielding \( \ln|t| + C \) once solved.Key integration rules beneficial here include:
  • Natural logarithm integral: \( \int \frac{1}{t} \, dt = \ln|t| + C \)
  • Recognizing forms that lead to standard integrals capable of easy evaluation
Such techniques eliminate complexities in integration processes, thus paving a simplified path towards outlining a comprehensive general solution to given linear differential equations. Mastering these techniques facilitates engaging with a broad array of more complex or non-standard integrals.

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Most popular questions from this chapter

Uniqueness is not just an abstraction designed to please theoretical mathematicians. For example, consider a cylindrical drum filled with water. A circular drain is opened at the bottom of the drum and the water is allowed to pour out. Imagine that you come upon the scene and witness an empty drum. You have no idea how long the drum has been empty. Is it possible for you to determine when the drum was full? (a) Using physical intuition only, sketch several possible graphs of the height of the water in the drum versus time. Be sure to mark the time that you appeared on the scene on your graph. (b) It is reasonable to expect that the speed at which the water leaves through the drain depends upon the height of the water in the drum. Indeed, Torricelli's law predicts that this speed is related to the height by the formula \(v^{2}=2 g h\), where \(g\) is the acceleration due to gravity near the surface of the earth. Let \(A\) and \(a\) represent the area of a cross section of the drum and drain, respectively. Argue that \(A \Delta=a v \Delta t\), and in the limit, \(A d h / d t=a v\). Show that \(d h / d t=-(a / A) \sqrt{2 g h}\). (c) By introducing the dimensionless variables \(\omega=\alpha h\) and \(s=\beta t\) and then choosing parameters $$ \alpha=\frac{1}{h_{0}} \quad \text { and } \quad \beta=\left(\frac{a}{A}\right) \sqrt{\frac{2 g}{h_{0}}}, $$ where \(h_{0}\) represents the height of a full tank, show that the equation \(d h / d t=-(a / A) \sqrt{2 g h}\) becomes \(d w / d s=-\sqrt{w}\). Note that when \(w=0\), the tank is empty, and when \(w=1\), the tank is full. (d) You come along at time \(s=s_{0}\) and note that the tank is empty. Show that the initial value problem, \(d w / d s=-\sqrt{w}\), where \(w\left(s_{0}\right)=0\), has an infinite number of solutions. Why doesn't this fact contradict the uniqueness theorem? Hint: The equation is separable and the graphs you drew in part (a) should provide the necessary hint on how to proceed.

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