Question

In Exercises \(16-24\) find the general solution. $$ x y^{\prime}+2 y=\frac{2}{x^{2}}+1 $$

Short answer

Answer: The general solution is \(y(x) = \frac{2}{x} + \frac{1}{2}x + \frac{C}{x^2}\).

Step-by-step solution

Rewrite the equation in standard form

Divide both sides of the equation by x:$$ y' + \frac{2}{x}y = \frac{2}{x^3} + \frac{1}{x} $$This gives the standard form of the first-order linear differential equation:$$ y' + P(x)y = Q(x) $$where \(P(x) = \frac{2}{x}\) and \(Q(x) = \frac{2}{x^3} + \frac{1}{x}\).

Determine the integrating factor

Calculate the integrating factor (\(\mu\)) by finding the exponential of the integral of \(P(x)\):$$ \mu(x) = e^{\int P(x)\,dx} = e^{\int \frac{2}{x}\,dx} $$Integrate \(\frac{2}{x}\):$$ \int \frac{2}{x}\,dx = 2 \int \frac{1}{x}\,dx = 2 \ln|x| + C $$The constant C does not matter when we multiply it by the differential equation, so the integrating factor is:$$ \mu(x) = e^{2\ln|x|} = x^2 $$

Multiply the equation by the integrating factor

Multiply both sides of the equation by the integrating factor \(x^2\):$$ x^2y' + 2xy = 2 + x $$This simplifies to:$$ (x^2y)' = 2 + x $$

Integrate both sides of the equation

Integrate both sides with respect to x:$$ \int (x^2y)' \, dx = \int (2 + x) \, dx $$This gives$$ x^2y = 2x + \frac{1}{2}x^2 + C $$

Isolate y to find the general solution

Solve for y by dividing both sides by \(x^2\), resulting in the general solution:$$ y(x) = \frac{2x + \frac{1}{2}x^2 + C}{x^2} = \frac{2}{x} + \frac{1}{2}x + \frac{C}{x^2} $$This is the general solution to the given differential equation.

Key concepts

First-Order Linear Differential Equation

Solving a first-order linear differential equation involves finding a function y(x) that satisfies the equation when substituted along with its derivatives. These equations take the form of \( y' + P(x)y = Q(x) \), where \( y' \) is the derivative of y with respect to x, \(P(x)\) is a function of x, and \(Q(x)\) represents the non-homogeneous part of the equation.

For instance, the differential equation from the exercise \( x y' + 2y = \frac{2}{x^2} + 1 \) was rewritten after dividing through by x to match this standard form, showcasing \(P(x)\) as \(\frac{2}{x}\) and \(Q(x)\) as \(\frac{2}{x^3} + \frac{1}{x}\). This allowed for the identification and application of an integrating factor to find the general solution, illustrating the characteristic steps taken to solve such equations.

Integrating Factor

An integrating factor is a strategic function used in the process of solving linear differential equations that simplifies the equation to an exact differential, making it easier to integrate.

To determine the integrating factor \( \mu(x) \), students compute \( e^{\int P(x)\,dx} \), with \(P(x)\) derived from the standard form of the equation. We found the integrating factor for the exercise to be \(x^2\) by integrating \(P(x) = \frac{2}{x}\) and utilizing exponent rules. By multiplying the entire differential equation by this factor, we created an easily integrable equation, \( (x^2y)' \), which paved the way for finding the solution. The integrating factor is a powerful tool in converting a complex-looking equation into a workable form.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They appear extensively in many fields such as physics, engineering, and economics to model real-world behaviors where change over time or space is considered.

There are several types of differential equations, and they are categorized based on order (the highest derivative involved) and linearity. The exercise outlined a first-order linear differential equation, indicating that the highest derivative involved is the first derivative, and the equation can be rearranged into a linear format. Besides first-order, equations can be of higher orders and may also be non-linear, requiring different approaches to solve. Understanding how to solve such equations provides a foundational tool for students in various scientific and engineering disciplines.

Boundary Value Problems

Boundary value problems (BVP) are specific types of differential equation problems where the solution is sought under certain fixed conditions, called boundary conditions. These conditions specify the values of the solution function on the boundary of the domain over which the differential equation is defined.

For example, if a differential equation is defined on a range of \( x \), [a, b], then boundary conditions might state what \( y(a) \) and \( y(b) \) should be. The exercise addressed did not involve a BVP as it sought the general solution of a differential equation without additional boundary conditions. However, applying boundary conditions to the general solution would yield a specific solution tailored to a particular scenario, which is often the case in practical applications where initial or end states are known.