Question

Solve the initial-value problems. $$ \frac{d y}{d x}+\frac{y}{2 x}=\frac{x}{y^{9}}, \quad y(1)=2. $$

Short answer

The particular solution of the given initial-value problem is: $$ x^{\frac{1}{2}}y = -\frac{1}{4}x^{\frac{1}{2}} + \int x^{\frac{3}{2}} y^{-9}dx + \frac{9}{4}. $$

Step-by-step solution

Identify the differential equation's form

The given differential equation is in the form of a first-order linear differential equation: $$ \frac{dy}{dx} + P(x)y = Q(x) $$ with \(P(x) = \frac{1}{2x}\) and \(Q(x) = \frac{x}{y^9}\).

Find the integrating factor

To find the integrating factor, we need to calculate the exponential of the integral of \(P(x)\): $$ \text{Integrating Factor} = e^{\int P(x) dx} = e^{\int \frac{1}{2x} dx}. $$ Now, we will calculate the integral: $$ \int \frac{1}{2x} dx = \frac{1}{2} \int \frac{1}{x} dx = \frac{1}{2} \ln(x) + C, $$ where \(C\) is the constant of integration. Now, we don't need to worry about the constant of integration as we are going to take its exponential to find the integrating factor, and any constant will not affect the result. Thus, $$ \text{Integrating Factor} = e^{\frac{1}{2}\ln(x)} = x^{\frac{1}{2}}. $$

Multiply the original differential equation by the integrating factor

Now, we will multiply the original differential equation by the integrating factor, \(x^{\frac{1}{2}}\): $$ x^{\frac{1}{2}}\left(\frac{dy}{dx} + \frac{1}{2x}y\right) = x^{\frac{1}{2}}\cdot\frac{x}{y^9}. $$ After multiplying, the equation becomes: $$ x^{\frac{1}{2}}\frac{dy}{dx} + \frac{1}{2}x^{-\frac{1}{2}}y = x^{\frac{1}{2}}\cdot\frac{x}{y^9}. $$

Integrate the exact differential equation

Now, we will integrate the obtained equation with respect to \(x\): $$ \int\left(x^{\frac{1}{2}}\frac{dy}{dx} + \frac{1}{2}x^{-\frac{1}{2}}y\right) dx = \int x^{\frac{1}{2}}\cdot\frac{x}{y^9}dx. $$ Integration of the left side: $$ \int x^{\frac{1}{2}} dy + \int \frac{1}{2}x^{-\frac{1}{2}}y dx = \int x^{\frac{3}{2}} y^{-9}dx. $$ Integrating each term, we get: $$ x^{\frac{1}{2}}y = -\frac{1}{4}x^{\frac{1}{2}} + \int x^{\frac{3}{2}} y^{-9}dx + C. $$ Here, we will consider the constant of integration (\(C\)) in the last term only since it will be added after integration, regardless of the term in which it is added.

Apply the initial condition

Now, we will apply the initial condition, \(y(1) = 2\), to find the particular solution: $$ (1)^{\frac{1}{2}}(2) = -\frac{1}{4}(1)^{\frac{1}{2}} + \int (1)^{\frac{3}{2}} (2)^{-9}d(1) + C. $$ Thus, $$ 2 = -\frac{1}{4} + C. $$ So, the constant \(C = \frac{9}{4}\).

Write the particular solution

Finally, the particular solution of the differential equation is: $$ x^{\frac{1}{2}}y = -\frac{1}{4}x^{\frac{1}{2}} + \int x^{\frac{3}{2}} y^{-9}dx + \frac{9}{4}. $$ This is the solution of the given initial-value problem.

Key concepts

First-order Linear Differential Equation

A first-order linear differential equation is a type of differential equation that involves first derivatives of a function. These equations typically have the form \(\frac{dy}{dx} + P(x)y = Q(x)\). This means the equation is linear in terms of the unknown function \(y\) and its derivative \(\frac{dy}{dx}\).

In our exercise, the function \(P(x)\) was \(\frac{1}{2x}\), and \(Q(x)\) is \(\frac{x}{y^9}\). The goal with these types of equations is often to find a function \(y(x)\) that satisfies the equation for all values within a certain range. To solve this equation type, identifying the right structure is crucial, which leads to using the right solving methods like the integrating factor.

Integrating Factor Method

The integrating factor method is a standard technique used to solve first-order linear differential equations. The idea is to simplify the differential equation by multiplying it by a specific function, known as the integrating factor, which allows the left-hand side of the equation to become an exact derivative. This combines the original equation into something that can be easily integrated.

To find the integrating factor in the given problem, we calculated the exponential of the integral of \(P(x)\), which is \(e^{\int \frac{1}{2x} \, dx}\). Solving this gives us \(x^{\frac{1}{2}}\). By multiplying the initial differential equation by this integrating factor, the equation simplifies greatly, making further solution steps possible.

This method transforms complex differential equations into ones that are easier to manage, often reducing them to simple integrals.

Particular Solution

A particular solution to a differential equation is a solution that satisfies both the differential equation and an initial or boundary condition. The initial condition is a specific requirement such as a given value of the function at a certain point. In our example, this was \(y(1) = 2\).

By applying the initial condition, we are able to find the exact form of the constant \(C\) involved in our integrated equation. From the exercise, after incorporating the initial condition, we calculated \(C = \frac{9}{4}\).

This process ensures that the solution we find is tailored specifically to the given problem, providing a unique solution that not only satisfies the differential equation generally but aligns perfectly with the initial conditions.

Differential Equations

Differential equations describe processes with rates of change. They involve an unknown function and its derivatives, symbolizing how a function evolves over time or space. These equations are categorized by order, type, and linearity.

First-order differential equations, like the one in the exercise, involve only the first derivative. They're foundational as they help model many phenomena in physics, biology, and engineering such as population growth and heat transfer.

Understanding the nature of a differential equation aids in determining the best approach for solving it. Recognizing whether it is linear or nonlinear can significantly impact the solving technique applied. Linear equations typically allow for systematic solution methods like the integrating factor method, whereas nonlinear ones may require numerical solutions or special integrating techniques.