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

(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5

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Rewrite the given differential equation in the standard linear form \(\displaystyle \frac{dy}{dx} + P(x)y = Q(x)\). Start by dividing both sides of the equation by \((x+1)\) to isolate \(\frac{dy}{dx}\):

\[\displaystyle \frac{dy}{dx} - \frac{2(x^2 + x)}{x+1} y = \frac{e^{x^2}}{(x+1)^2}.\]

Simplify the coefficient of \(y\) by factoring and reducing the expression \(\frac{2(x^2 + x)}{x+1}\). Notice that \(x^2 + x = x(x+1)\), so this becomes \(\frac{2x(x+1)}{x+1} = 2x\) for \(x > -1\).

Now the equation is \(\displaystyle \frac{dy}{dx} - 2x y = \frac{e^{x^2}}{(x+1)^2}\). Identify \(P(x) = -2x\) and \(Q(x) = \frac{e^{x^2}}{(x+1)^2}\).

Find the integrating factor \(\mu(x)\) using the formula \(\displaystyle \mu(x) = e^{\int P(x) dx} = e^{\int -2x dx} = e^{-x^2}\). Multiply both sides of the differential equation by \(\mu(x)\) to write the left side as the derivative of a product:

\[\displaystyle e^{-x^2} \frac{dy}{dx} - 2x e^{-x^2} y = \frac{e^{x^2}}{(x+1)^2} e^{-x^2} \implies \frac{d}{dx} \left( y e^{-x^2} \right) = \frac{1}{(x+1)^2}.\]

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주요 개념

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

Initial Value Problems (IVPs)

An initial value problem involves solving a differential equation with a given initial condition, which specifies the value of the unknown function at a particular point. This condition allows for finding a unique solution that fits both the differential equation and the initial value.

First-Order Linear Differential Equations

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

Integrating Factor Method

The integrating factor is a function, usually denoted μ(x), used to multiply a linear differential equation to make the left side an exact derivative. It is typically μ(x) = exp(∫P(x) dx), enabling straightforward integration and solution of the equation.

Solving Initial Value ProblemsSolve the initial value problems in Exercises 15–20.(x+1) dy/dx - 2 (x² + x)y = exp(x²) / (x+1), x > -1, y(0) = 5