Math Problem Statement

Find the general solution of the given differential equation, and use it to determine how solutions behave as t → ∞. 8y' + y = 9t^2

Solution

The differential equation provided is:

8y+y=9t28y' + y = 9t^2

To solve this, we will use the method for solving first-order linear differential equations.

Step 1: Rewrite the equation

We can rewrite the equation in standard form:

y+18y=98t2y' + \frac{1}{8}y = \frac{9}{8}t^2

Step 2: Find the integrating factor

The integrating factor, μ(t)\mu(t), is given by:

μ(t)=e18dt=et8\mu(t) = e^{\int \frac{1}{8} \, dt} = e^{\frac{t}{8}}

Step 3: Multiply through by the integrating factor

Multiplying both sides of the equation by et8e^{\frac{t}{8}}:

et8y+18et8y=98t2et8e^{\frac{t}{8}}y' + \frac{1}{8}e^{\frac{t}{8}}y = \frac{9}{8}t^2e^{\frac{t}{8}}

The left side is now the derivative of yet8y \cdot e^{\frac{t}{8}}, so we can rewrite the equation as:

ddt(yet8)=98t2et8\frac{d}{dt} \left( y \cdot e^{\frac{t}{8}} \right) = \frac{9}{8}t^2e^{\frac{t}{8}}

Step 4: Integrate both sides

We now integrate both sides:

yet8=98t2et8dty \cdot e^{\frac{t}{8}} = \int \frac{9}{8}t^2e^{\frac{t}{8}} \, dt

We can solve the right-hand side using integration by parts. The result of the integration is:

yet8=9(t216(t8))+Cy \cdot e^{\frac{t}{8}} = 9 \left( t^2 - 16(t - 8) \right) + C

where CC is the constant of integration. Dividing both sides by et8e^{\frac{t}{8}}, we get:

y=9(t216(t8))+Cet8y = 9 \left( t^2 - 16(t - 8) \right) + \frac{C}{e^{\frac{t}{8}}}

Step 5: Behavior as tt \to \infty

As tt \to \infty, the term involving C/et8C/e^{\frac{t}{8}} approaches zero because the exponential in the denominator grows very quickly. Therefore, the solution converges to:

y=9(t216(t8))y = 9 \left( t^2 - 16(t - 8) \right)

Thus, the function to which the solution converges is:

y=9(t216(t8))y = 9 \left( t^2 - 16(t - 8) \right)

Let me know if you'd like more details or clarifications on any step!

Follow-up Questions:

  1. What is the purpose of the integrating factor in solving first-order differential equations?
  2. How do we know when to use the method of integrating factors?
  3. What other methods could be used to solve this type of differential equation?
  4. How does the solution change if the differential equation includes additional terms (e.g., yy terms)?
  5. How can we interpret the physical meaning of the solution for large values of tt?

Tip:

When solving differential equations, always check if the equation is separable, linear, or exact, as the solution method depends on the structure of the equation.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
First-Order Linear Equations
Integrating Factors

Formulas

Standard form of first-order linear differential equation: y' + P(t)y = Q(t)
Integrating factor: e^(∫P(t) dt)

Theorems

Method of Integrating Factors

Suitable Grade Level

Undergraduate Level