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Magazine Chapter 2: Problem 44
Solve the given differential equation subject to the indicated initial condition. $$ \frac{d y}{d t}+t y=y, \quad y(1)=3 $$
Short Answer
Expert verified
The solution is \( y(t) = 3 e^{-(t^2/2 - t - 1/2)} \).
Step by step solution
01
Simplify the Differential Equation
Start by rewriting the equation \( \frac{d y}{d t} + t y = y \) as \( \frac{d y}{d t} + t y - y = 0 \). This simplifies to \( \frac{d y}{d t} + (t-1) y = 0 \). This is a first-order linear homogeneous differential equation.
02
Identify Standard Linear Form
The standard form of a first-order linear differential equation is \( \frac{d y}{d t} + P(t)y = Q(t) \). Here, \( P(t) = t-1 \) and \( Q(t) = 0 \).
03
Integrating Factor
Calculate the integrating factor \( \mu(t) \) using \( e^{\int P(t) \, dt} \). Since \( P(t) = t-1 \), find the integrating factor: \[ \mu(t) = e^{\int (t-1) \, dt} = e^{t^2/2 - t} \].
04
Transform the Equation
Multiply the entire differential equation by the integrating factor \( \mu(t) \) to obtain: \( e^{t^2/2 - t} \frac{d y}{d t} + (t-1) e^{t^2/2 - t} y = 0 \).
05
Integrate to Solve
Since the left-hand side is the derivative of \( y \cdot \mu(t) \), integrate both sides: \[ \int \frac{d}{d t} (y \cdot e^{t^2/2 - t}) \ dt = \int 0 \, dt \]. This simplifies to: \( y \cdot e^{t^2/2 - t} = C \).
06
Solve for y
Solve for \( y \) by dividing both sides by \( e^{t^2/2 - t} \): \( y = C \, e^{-(t^2/2 - t)} \).
07
Apply Initial Conditions
Use the initial condition \( y(1) = 3 \) to find \( C \). Substitute \( t = 1 \) and \( y = 3 \) into the equation: \[ 3 = C \, e^{-(1^2/2 - 1)} = C \, e^{1/2} \]. Solve for \( C \): \( C = 3e^{1/2} \).
08
Final Solution
Substitute \( C = 3e^{1/2} \) back into the equation for \( y \): \[ y(t) = 3 e^{1/2} e^{-(t^2/2 - t)} \]. Simplify to: \( y(t) = 3 e^{-(t^2/2 - t - 1/2)} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Conditions
Initial conditions are specific values given for the variables in a differential equation that allow us to find a particular solution that satisfies these conditions. In this exercise, the initial condition is specified as \( y(1) = 3 \). This means that when the independent variable \( t \) is 1, the dependent variable \( y \) is 3.
- Initial conditions help in determining the constant of integration, \( C \), which differentiates a general solution from a particular one.
- They are crucial because, without them, you can only solve differential equations up to a constant.
- Once \( C \) is determined with help from the initial conditions, you can substitute it back into the general solution to get a specific result.
Integrating Factor
The integrating factor is a useful tool for solving first-order linear differential equations. It transforms a differential equation into one that is easier to integrate. For a given equation of the form \( \frac{d y}{d t} + P(t) y = Q(t) \), the integrating factor, \( \mu(t) \), is calculated using the expression:
\[\mu(t) = e^{\int P(t) \, dt} \]
\[\mu(t) = e^{\int P(t) \, dt} \]
- This function, when multiplied with the differential equation, helps in simplifying the integration process by making the left side a straightforward derivative.
- After transformation, the equation becomes \( \frac{d}{d t} (y \cdot \mu(t)) = Q(t) \cdot \mu(t) \), where the left-hand side is now the derivative of a product.
Homogeneous Differential Equation
A homogeneous differential equation has the form where the dependent variable and its derivatives are set to zero. In this scenario, the differential equation \( \frac{d y}{d t} + (t-1) y = 0 \) is homogeneous.
- Homogeneous equations often appear when the regular components are initially organized and simplified to zero after transformation.
- These equations have particular characteristics suitable for using methods like integrating factors because it simplifies to simple derivative steps.
Solution Method
The solution method for a first-order linear differential equation involves several clear steps similar to a recipe. Here鈥檚 the breakdown:
- Simplify the Equation: First, rearrange the terms if necessary to fit the standard form \( \frac{d y}{d t} + P(t)y = Q(t) \).
- Identify \( P(t) \) and \( Q(t) \): In our example, \( P(t) = t-1 \) and \( Q(t) = 0 \).
- Calculate the Integrating Factor: Use the expression \( \mu(t) = e^{\int P(t) \, dt} \) which becomes \( e^{t^2/2 - t} \).
- Transform and Solve: Multiply the whole differential equation by the integrating factor to get the derivative of a product and integrate to solve for \( y \cdot \mu(t) \).
- Solve for \( y \): Finally, isolate \( y \) and apply any initial condition to find the constant of integration, \( C \).
