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Magazine Chapter 8: Problem 19
Solve the differential equation. $$ y^{\prime}+y=e^{-t} \ln t $$
Short Answer
Expert verified
The solution is \( y = \frac{t \ln t - t + C}{e^{t}} \).
Step by step solution
01
Identify the type of differential equation
The given differential equation is a linear first-order differential equation of the form \[ y' + p(t)y = g(t) \] where \[ p(t) = 1 \] and \[ g(t) = e^{-t} \, \ln t \]
02
Find the integrating factor
The integrating factor \( \mu(t) \) is found using the formula: \[ \mu(t) = e^{\int p(t) \, dt} = e^{\int 1 \, dt} = e^{t} \]
03
Multiply through by the integrating factor
Multiply every term in the differential equation by the integrating factor \( e^{t} \): \[ e^{t}y' + e^{t}y = e^{t} \, e^{-t} \, \ln t \] which simplifies to \[ e^{t}y' + e^{t}y = \ln t \]
04
Rewrite the left side as a product rule
The left side of the equation can be written as the derivative of \( e^{t}y \): \[ \frac{d}{dt}(e^{t}y) = \ln t \]
05
Integrate both sides
Integrate both sides with respect to \( t \): \[ \int \frac{d}{dt}(e^{t}y) \, dt = \int \ln t \, dt \] This gives: \[ e^{t}y = \int \ln t \, dt \]
06
Evaluate the integral on the right side
Use integration by parts for \( \int \ln t \, dt \): Let \( u = \ln t \) and \( dv = dt \). Then, \( du = \frac{1}{t}dt \) and \( v = t \). Using the integration by parts formula, \( \int u \, dv = uv - \int v \, du \), we get: \[ \int \ln t \, dt = t \ln t - \int t \cdot \frac{1}{t} \, dt = t \ln t - t + C \]
07
Solve for y
Now we have: \[ e^{t}y = t \ln t - t + C \] Solve for \( y \) by dividing both sides by \( e^{t} \): \[ y = \frac{t \ln t - t + C}{e^{t}} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
linear first-order differential equation
When you encounter a differential equation, the first thing to understand is its type. In this case, we are dealing with a linear first-order differential equation. This type has the general form: \( y' + p(t)y = g(t) \). Here, \( y' \) is the first derivative of \( y \) with respect to time \( t \). This form consists of three main parts:
- \(y'\) - the derivative of \(y\)
- \(p(t)\) - a function that is multiplied by \(y\)
- \(g(t)\) - a non-homogeneous term, which is a function of \( t \)
integrating factor
To solve a linear first-order differential equation, we use an integrating factor. This makes the equation easier to solve. The integrating factor for our form \( y' + p(t)y = g(t) \) is given by: \[ \mu\(t\) = e^{\int p(t) \, dt} \]. Here's a step-by-step process:
- Calculate the integral of \( p(t) \).
- Exponentiate the result to find the integrating factor.
integration by parts
At a certain point, we need to solve an integral, in this case, \( \int \ln t \, dt \). To handle this, we use an advanced technique called integration by parts. The integration by parts formula is: \[ \int u \, dv = uv - \int v \, du \]. Here鈥檚 how to apply it:
- Choose \(u\) and \(dv\) from the integral. Generally, let \(u\) be the part that becomes simpler when differentiated.
- Differentiating and integrating parts separately, getting \(du\) and \(v\).
- Plug the parts into the formula to find the integral.
- Let \( u = \ln t \) and \( dv = dt \).
- Then, \( du = \frac{1}{t}dt \) and \( v = t \).
product rule
The product rule is a handy tool for derivatives. It states that the derivative of a product of two functions is given by: \( (fg)' = f'g + fg' \). This is useful in our differential equation problem. When we multiply by the integrating factor \( e^{t} \), the left-hand side becomes the product rule of \( e^{t}y \):
- The equation \( e^{t}y' + e^{t}y = (e^{t}y)' \).
