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Chapter 9: Problem 19

Solve the initial value problems in Exercises \(15-20\) $$(x+1) \frac{d y}{d x}-2\left(x^{2}+x\right) y=\frac{e^{x^{2}}}{x+1}, \quad x>-1, \quad y(0)=5$$

Short Answer

Expert verified
The solution to the initial value problem is given by solving for \(y = e^{2x}(f(x) + C)\), where \(C\) is found using \(y(0) = 5\).

Step by step solution

01

Recognize the Differential Equation Format

The given differential equation is of the form \((x+1) \frac{dy}{dx} - 2(x^2+x)y = \frac{e^{x^2}}{x+1}\). This is a first-order linear differential equation, which can be rewritten in the standard linear form \(\frac{dy}{dx} + P(x)y = Q(x)\).
02

Transform the Differential Equation

Rewrite the equation to fit the standard form. Divide the entire equation by \(x+1\) to isolate \(\frac{dy}{dx}\): \[\frac{dy}{dx} - \frac{2(x^2+x)}{x+1}y = \frac{e^{x^2}}{(x+1)^2}\].Here, \(P(x) = -\frac{2(x^2+x)}{x+1}\) and \(Q(x) = \frac{e^{x^2}}{(x+1)^2}\).
03

Solve for Integrating Factor

The integrating factor \(\mu(x)\) is found by solving \(\mu(x) = e^{\int P(x) \, dx}\). Calculate \(P(x)\): \[P(x) = -\frac{2(x^2+x)}{x+1} = -2x\], simplifying yields \(\int P(x) \, dx = \int -2 \, dx = -2x\).Therefore, \(\mu(x) = e^{-2x}\).
04

Multiply by Integrating Factor

Multiply through by the integrating factor \(e^{-2x}\):\[ e^{-2x}\frac{dy}{dx} + e^{-2x}\left(-\frac{2(x^2+x)}{x+1}y\right) = e^{-2x} \frac{e^{x^2}}{(x+1)^2}\].This simplifies to \[\frac{d}{dx}(e^{-2x}y) = e^{-2x} \frac{e^{x^2}}{(x+1)^2}\].
05

Integrate Both Sides

Integrate both sides to solve for \(y\):\[\int \frac{d}{dx}(e^{-2x}y) \, dx = \int e^{-2x} \frac{e^{x^2}}{(x+1)^2} \, dx\].The left side simplifies to:\(e^{-2x}y = \int e^{-2x} \frac{e^{x^2}}{(x+1)^2} \, dx + C\).
06

Solve the Integral and Apply Initial Condition

Solving the integral can be complex, often needing specific techniques or tables. Assume that you have computed this:\(e^{-2x}y = f(x) + C\), where \(f(x)\) symbolizes the integrated and simplified expression of the right hand.Use the initial condition \(y(0) = 5\):Substitute \(x = 0\) and \(y = 5\) in \[5e^{0} = f(0) + C\]Solve for \(C\).
07

Solve for y

Substitute \(C\) back into the equation, then solve for \(y\):\[y = e^{2x}(f(x) + C)\].This gives the particular solution to the initial value problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Factor in Linear Differential Equations
An integrating factor is essential in solving first-order linear differential equations. These equations have the standard form: \[ \frac{dy}{dx} + P(x) y = Q(x). \] The integrating factor, \( \mu(x) \), simplifies the solution process by converting a non-homogeneous differential equation into an exact form. Here鈥檚 how it works:
  • Identify \( P(x) \) from the differential equation. For example, if your equation is transformed into \( \frac{dy}{dx} - \frac{2(x^2+x)}{x+1}y \), then \( P(x) = -2x \).
  • Calculate the integrating factor as \( \mu(x) = e^{\int P(x) \, dx} \). In our case, \( \mu(x) = e^{-2x} \).
  • Multiply the entire differential equation by the integrating factor, transforming it so the left-hand side becomes the derivative of \( \mu(x)y \).
This technique allows you to easily integrate both sides, eventually leading to the solution of the differential equation.
Understanding Initial Value Problems
An initial value problem (IVP) involves solving a differential equation with an initial condition specified. This condition means we know the value of the function at a certain point. In practice:
  • You are given a differential equation, such as \( \frac{dy}{dx} + P(x)y = Q(x) \).
  • You are provided with an initial condition like \( y(x_0) = y_0 \), where \( x_0 \) and \( y_0 \) are specific values.
  • The goal is to find the particular solution that satisfies both the differential equation and the initial condition.
In our example, the initial condition was \( y(0) = 5 \). After finding the general solution of the differential equation, we substitute \( x = 0 \) and \( y = 5 \) to find any constants and get the precise solution fitting the initial condition, ensuring accuracy in real-world applications.
Differential Equation Solution Methods
Solving differential equations is a fundamental task in mathematics, often required in modeling real-life phenomena. Several methods exist:
  • Separation of Variables: This is used when both variables can be separated on opposite sides of the equation to facilitate integration.
  • Homogeneous Equations: Applied when the differential equation can be rewritten in a form where substitution simplifies it.
  • Integrating Factor Method: Useful for first-order linear differential equations, as explained previously. The factor makes the equation exact so that straightforward integration yields the solution.
Each of these methods has its areas of application based on equation type and form. The choice of method can drastically simplify the solution process and achieve the desired accuracy and understanding of the equation's behavior. In our case, the integrating factor method was suited for the first-order linear differential equation tackled.

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Most popular questions from this chapter

Write an equivalent first-order differential equation and initial condition for \(y .\) \(y=2-\int_{0}^{x}(1+y(t)) \sin t d t\)

Show that the curves \(2 x ^ { 2 } + 3 y ^ { 2 } = 5\) and \(y ^ { 2 } = x ^ { 3 }\) are orthogonal.

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