Solving Initial Value Problems
Solve the initial value problems in Exercises 15–20.

t dy/dt + 2y = t³, t > 0, y(2) = 1

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Rewrite the given differential equation in standard linear form by dividing both sides by \(t\) (since \(t > 0\)): \[\frac{dy}{dt} + \frac{2}{t} y = t^2\]

Identify the integrating factor \(\mu(t)\), which is given by: \[\mu(t) = e^{\int P(t) \, dt} = e^{\int \frac{2}{t} \, dt}\]

Calculate the integrating factor \(\mu(t)\) explicitly by evaluating the integral: \[\mu(t) = e^{2 \ln |t|} = e^{\ln |t|^2} = t^2\] (since \(t > 0\), absolute value can be dropped)

Multiply both sides of the differential equation by the integrating factor \(t^2\) to write the left side as a derivative of a product: \[t^2 \frac{dy}{dt} + 2t y = t^4\] which can be expressed as: \[\frac{d}{dt} (t^2 y) = t^4\]

Integrate both sides with respect to \(t\): \[\int \frac{d}{dt} (t^2 y) \, dt = \int t^4 \, dt\] This gives: \[t^2 y = \frac{t^5}{5} + C\] where \(C\) is the constant of integration. Use the initial condition \(y(2) = 1\) to solve for \(C\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

First-Order Linear Differential Equations

These are differential equations of the form dy/dt + P(t)y = Q(t). They can be solved using an integrating factor, which simplifies the equation into an exact derivative, allowing integration on both sides to find the general solution.

Integrating Factor Method

The integrating factor is a function, usually denoted μ(t) = e^(∫P(t) dt), used to multiply both sides of a linear differential equation. This transforms the left side into the derivative of (μ(t)y), making it easier to integrate and solve for y.

Initial Value Problems (IVP)

An IVP specifies the value of the unknown function at a particular point, such as y(2) = 1. After finding the general solution, the initial condition is used to determine the specific constant, yielding a unique solution to the differential equation.

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.t dy/dt + 2y = t³, t > 0, y(2) = 1