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Chapter 10: Problem 22

Solve the initial-value problem. $$t y^{\prime}+y=\ln t, y(e)=0, t>0$$

Short Answer

Expert verified
The solution is \( y = \ln t - 1 \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is a first-order linear ordinary differential equation of the form: \[t y^{\prime} + y = \ln t\]
02

Rewrite the Equation in Standard Form

The standard form for a linear differential equation is: \[y^{\backslash prime} + P(t)y = Q(t)\]Divide every term by \( t \) to obtain:\[y^{\backslash prime} + \frac{1}{t} y = \frac{\ln t}{t}\]
03

Determine the Integrating Factor

Identify \(P(t) = \frac{1}{t}\). The integrating factor \( \mu(t) \) is found using:\[ \mu(t) = e^\int P(t) dt = e^\int \frac{1}{t} dt = e^\ln t = t \]
04

Multiply the Differential Equation by the Integrating Factor

Multiply every term in the equation \[y^{\backslash prime} + \frac{1}{t} y = \frac{\ln t}{t}\] by the integrating factor \( t \):\[ t y^{\backslash prime} + y = (\ln t)\]
05

Integrate Both Sides

Rewrite the left side as a derivative of a product: \[ \frac{d}{dt}(t y) = \ln t \]Integrate both sides with respect to \( t \):\[ t y = \int \ln t \, dt \]
06

Compute the Integral

To integrate \( \int \ln t \, dt \), use integration by parts with \( u = \ln t \) and \( dv = dt \). Then \( du = \frac{1}{t} dt \) and \( v = t \).The integral \( \int \ln t \, dt \) becomes:\[ t y = t \ln t - \int t \cdot \frac{1}{t} dt = t \ln t - \int 1 dt = t \ln t - t + C \]
07

Solve for y

Divide both sides by \( t \) to solve for \( y \):\[ y = \ln t - 1 + \frac{C}{t} \]
08

Use the Initial Condition

Apply the initial condition \( y(e) = 0 \):\[ 0 = \ln e - 1 + \frac{C}{e} \]Since \( \ln e = 1 \), this simplifies to:\[ 0 = 1 - 1 + \frac{C}{e} \Rightarrow \frac{C}{e} = 0 \Rightarrow C = 0 \]
09

Write the Final Solution

Since \( C = 0 \), the final solution is:\[ y = \ln t - 1 \]

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Key Concepts

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

First-Order Linear Ordinary Differential Equation
A first-order linear ordinary differential equation (ODE) is an equation of the form:
  • \(\frac{dy}{dt} + P(t)y = Q(t)\)
Here, \(y\) is the function we need to solve for, and \(P(t)\) and \(Q(t)\) are functions of \(t\).
The given equation, \(t y^{\backslash prime}+y=\text{ln}(t)\), is a first-order linear ODE.
It can be converted to its standard form by dividing through by \(t\), giving \(y^{\backslash prime} + \frac{1}{t} y = \frac{\text{ln}(t)}{t}\).
  • This makes it easier to identify \(P(t)\) and \(Q(t)\).
  • In this case: \(P(t) = \frac{1}{t} \) and \(Q(t) = \frac{\text{ln}(t)}{t} \).
Integrating Factor
The integrating factor is a crucial tool for solving first-order linear ODEs. It transforms the left-hand side of the differential equation into the derivative of a product of two functions.
To find the integrating factor \( \mu(t)\), we use the formula:
  • \( \mu(t) = e^{\int P(t) dt} \)
Given \( P(t) = \frac{1}{t}\), we compute:
\( \mu(t) = e^{\int \frac{1}{t} dt} = e^{\text{ln}(t)} = t \)
This integrating factor helps simplify the equation:
  • Multiply the original standard form equation by \( t \).
  • The left side becomes \(t y^{\backslash prime} + y\), which can be written as the derivative of \(y t\).
Integration by Parts
Integration by parts is a technique used to integrate products of functions. It follows from the product rule for differentiation. The formula is:
  • \(\backslash \text{int}u \ dv = uv - \backslash \text{int}v \ du\)
For our problem, to integrate \( \backslash \text{int} \text{ln}(t) \ dt \), we use:
  • \( u = \text{ln}(t) \)
  • \( dv = dt \)
Then, we find:
  • \( du = \frac{1}{t} dt \)

  • \( v = t \)
Thus, \( \backslash \text{int} \text{ln}(t) \ dt \ = t \text{ln}(t) - \backslash \text{int} t \left(\frac{1}{t}\right) dt = t \text{ln}(t) - t + C \).

Using this, the equation simplifies to:
  • \( \backslash y t = t \text{ln}(t) - t + C \)
  • Finally, solving for \( y \), we obtain the general solution: \( y = \text{ln} t - 1 + \frac{C}{t} \).

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Most popular questions from this chapter

Draw the graph of \(g(x)=e^{x}-100 x^{2}-1,\) and use the graph to sketch the solution of the differential equation \(y^{\prime}=e^{y}-100 y^{2}-1\) with initial condition \(y(0)=4\) on a ty-coordinate system.

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