The given differential equation is:

$x' = \frac{x}{t} - 2$

This is a first-order linear differential equation of the form:

$x' + P(t)x = Q(t)$

Where:

• $P(t) = -\frac{1}{t}$
• $Q(t) = -2$

We can solve this equation step-by-step as follows:

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Step 1: Write the standard form

The equation is already in standard form: $x' - \frac{1}{t}x = -2$

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Step 2: Find the integrating factor

The integrating factor $\mu(t)$ is given by: $\mu(t) = e^{\int P(t) \, dt}$ Substitute $P(t) = -\frac{1}{t}$: $\mu(t) = e^{\int -\frac{1}{t} \, dt} = e^{-\ln|t|} = \frac{1}{t}$

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Step 3: Multiply through by the integrating factor

Multiply the entire equation by $\frac{1}{t}$: $\frac{1}{t}x' - \frac{1}{t^2}x = -\frac{2}{t}$

The left-hand side becomes the derivative of $\frac{x}{t}$: $\frac{d}{dt}\left(\frac{x}{t}\right) = -\frac{2}{t}$

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Step 4: Integrate both sides

Integrate both sides with respect to $t$: $\frac{x}{t} = \int -\frac{2}{t} \, dt$

The integral of $-\frac{2}{t}$ is: $\int -\frac{2}{t} \, dt = -2 \ln|t| + C$

So: $\frac{x}{t} = -2 \ln|t| + C$

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Step 5: Solve for $x$

Multiply through by $t$ to isolate $x$: $x = t(-2 \ln|t| + C)$

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Final Solution:

$x(t) = -2t \ln|t| + Ct$

Where $C$ is the constant of integration.

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Do you want a detailed explanation of any specific step, or have any questions?

Related Questions:

1. What is an integrating factor, and how does it simplify linear differential equations?
2. How do logarithmic functions appear in solutions to differential equations?
3. What conditions would make this solution invalid (e.g., $t = 0$)?
4. Can we graph this solution for different values of $C$?
5. How do we interpret this solution physically, if $x$ and $t$ represent real-world quantities?

Tip: Always check the domain of the solution, especially when logarithmic terms like $\ln|t|$ are involved!