Question

In Exercises \(30-37\) solve the initial value problem. $$ (x+2) y^{\prime}+4 y=\frac{1+2 x^{2}}{x(x+2)^{3}}, \quad y(-1)=2 $$

Short answer

Question: Find the solution to the initial value problem \((x+2) y'+4y = \frac{1+2x^2}{x(x+2)^3}\) with the initial condition \(y(-1) = 2\). Answer: The solution to the initial value problem is given by: $$ y(x) = \frac{1}{(x+2)^4}\left(\int \frac{1+2x^2}{x}(x+2)dx + 2\right) $$

Step-by-step solution

Find the integrating factor

For a first-order linear differential equation \((x+2) y'+4y = F(x)\), the integrating factor is given by \(I(x) = e^{\int P(x) \, dx}\), with \(P(x)=\frac{4}{x+2}\). Let's find \(I(x)\): $$ I(x) = e^{\int \frac{4}{x+2} \, dx} = e^{4\int \frac{1}{x+2} \, dx} $$ Using substitution, let \(u = x+2\), so \(du = dx\). Therefore, we have: $$ I(x) = e^{4 \int \frac{1}{u}\, du} = e^{4 \ln(u)} = u^4 = (x+2)^4 $$ So the integrating factor, \(I(x) = (x+2)^4\).

Multiply the given equation by the integrating factor

Now, multiply the entire original equation by \(I(x)\): $$ (x+2)^4((x+2) y'+4y) = (x+2)^4\left(\frac{1+2x^2}{x(x+2)^3}\right) $$ This simplifies to: $$ (x+2) y' (x+2)^4 + 4y(x+2)^4 = \frac{1+2x^2}{x} (x+2) $$

Integrate the equation obtained in Step 2

Now, let's integrate both sides with respect to \(x\). Note that the left side is the derivative of the product \(y(x)(x+2)^4\), so we have: $$ \int \frac{d}{dx}[y(x)(x+2)^4]dx = \int \frac{1+2x^2}{x}(x+2)dx $$ After integrating, we have: $$ y(x)(x+2)^4 = \int \frac{1+2x^2}{x}(x+2)dx + C $$

Apply the initial condition to determine the constant of integration

Now, use the initial condition, \(y(-1) = 2\), to determine the constant of integration \(C\): $$ 2((-1)+2)^4 = \int \frac{1+2(-1)^2}{-1}((-1)+2) d(-1) + C $$ $$ 2(1)^4 = 0 + C $$ $$ C = 2 $$ So the constant of integration is \(C = 2\).

Write the final solution

Finally, we can write the solution to the initial value problem: $$ y(x)(x+2)^4 = \int \frac{1+2x^2}{x}(x+2)dx + 2 $$ Divide both sides by \((x+2)^4\) to solve for \(y(x)\): $$ y(x) = \frac{1}{(x+2)^4}\left(\int \frac{1+2x^2}{x}(x+2)dx + 2\right) $$ This is the final solution to the initial value problem.

Key concepts

First-Order Linear Differential Equation

A first-order linear differential equation is a type of equation that appears quite frequently in various fields of science and engineering. These equations contain a first derivative and are characterized by a specific structure. The standard form of a first-order linear differential equation is:
\[ a(x)y' + b(x)y = F(x) \]Here, \( y' \) is the derivative of \( y \) with respect to \( x \), \( a(x) \), \( b(x) \), and \( F(x) \) are given functions of \( x \). In the exercise provided, this equation takes the form: \((x+2) y'+4y = \frac{1+2x^2}{x(x+2)^3}\).
The goal is to find a function \( y(x) \) that satisfies this equation for all \( x \). One efficient way to tackle these problems is by using an integrating factor, which simplifies the equation and makes it easier to solve.

Integrating Factor

The integrating factor is a clever mathematical tool used to solve first-order linear differential equations. The main idea is to multiply the entire differential equation by a special function, the integrating factor, to transform it into a form that is easier to solve.
The process involves computing the integrating factor \( I(x) \) using the formula:

• \( I(x) = e^{\int P(x) \, dx} \)
• Here, \( P(x) = \frac{b(x)}{a(x)} \)

For the exercise, the integrating factor becomes \( I(x) = (x+2)^4 \). Multiplying the whole differential equation by this factor changes the left-hand side into the derivative of a product, which simplifies the integration process. By finding this factor, you essentially convert a complex differential equation into a simpler problem of integration.

Constant of Integration

When solving differential equations, especially when doing integration, it's necessary to find a constant of integration. This constant, denoted by \( C \), appears because indefinite integration yields a family of solutions instead of a single solution. The constant ensures all possible solutions can be accounted for.
In initial value problems, there is often additional information about the function, such as \( y(-1) = 2 \), which helps determine the exact value of \( C \). For our exercise, after integrating each side of the equation, the constant of integration was calculated to be \( C = 2 \) using the initial condition. This exact value allows us to determine the specific solution that fits the given initial condition.

Differential Equation Solution

A differential equation solution involves finding a function \( y(x) \) that satisfies the equation across its domain and meets the initial conditions if provided. Solving these equations often requires integrating parts of the equation once simplified by the integrating factor.
After finding the integrating factor and integrating both sides of the equation, we obtained:\[ y(x)(x+2)^4 = \int \frac{1+2x^2}{x}(x+2)dx + 2 \]Finally, dividing through by \((x+2)^4\) yields the final expression for \( y(x) \). This process encapsulates the full solution, providing a function that explains how \( y \) behaves in relation to \( x \). The ability to solve such equations is fundamental, as it provides insights into various phenomena described by differential equations.