Question

Finding a Particular Solution In Exercises \(17-24,\) find the particular solution of the differential equation that satisfies the initial condition. $$ \text{Differential Equation} \quad \text{Initial Condition} $$ $$ 2 x y^{\prime}-y=x^{3}-x \quad y(4)=2 $$

Short answer

The particular solution for the given differential equation with the initial condition \(y(4) = 2\) is \(y = (1/3)x^{3} - (1/2)x^{2} + (1/4)x\).

Step-by-step solution

Rewrite the Differential Equation into Standard Form

Rewrite the given differential equation into standard form by dividing each term by \(2x\). This results in \(y'-(1/2x)y = (1/2)x^{2}-(1/2x)\).

Solve the Homogeneous Equation

Solve for the homogeneous equation \(y'-(1/2x)y = 0\) using separation of variables by integrating \(dy/y = (1/2x)dx\), giving the homogeneous/general solution \(y=Ae^{ln|x|}\), where \(A\) is a constant.

Find a Particular Solution

Use the method of variation of parameters to find a particular solution, or integrating factor method to get \(e^{-ln|x|}\), and then multiply each term in equation by this factor, resulting in \(xy' - y = x^{2}-x\). We then integrate each side to get the general solution \(y = (1/3)x^{3} - (1/2)x^{2} + cx\), where \(c\) is a constant.

Use Initial Condition to Find the Constant

When \(x = 4\) and \(y = 2\), substituting into the general equation above gives \(2 = (1/3)(4)^{3} - (1/2)(4)^{2} + 4c\), when simplified gives \(c = 1/4\).

Key concepts

Initial Conditions

Initial conditions are crucial in the study of differential equations. They are specific conditions given at a point in time that help uniquely determine a particular solution from a family of possible solutions.
In our exercise, the initial condition is given as \( y(4) = 2 \), which means when \( x = 4 \), the value of \( y \) is 2.
This initial condition serves a vital role in finding the unique solution that satisfies not only the differential equation but also the given point in space. Here's why initial conditions matter:

• They help specify a particular solution from a general solution, making it unique.
• They allow modeling real-world scenarios where initial states are known.

This is much like tracing a specific path from an infinite number of paths generated by the same equation.

Homogeneous Equation

A homogeneous equation is a type of differential equation which can be expressed in the form \( y' + P(x)y = 0 \). It has no independent term, making its setup quite simple.
In our case, it's \( y' - (1/2x)y = 0 \). This involves separating the variables, making it solvable by integration.
The beauty of homogeneous equations is:

• They simplify into a form that can be directly integrated.
• The solutions often describe exponential growth or decay relationships due to their structure.

For our homogeneous equation, solving it gives \( y = Ae^{ ext{ln}|x|} \), where \( A \) is an arbitrary constant. This showcases how dividing through by \( y \) and separating the variables laid out a tidy path to an elegant solution.

Variation of Parameters

The method of variation of parameters is an invaluable technique for finding a particular solution to non-homogeneous differential equations.
Unlike the homogeneous solution, which applies to the simplified version of the problem, this method accounts for external, non-uniform influences. Here it helps solve problems like \( xy' - y = x^3 - x \).
Key aspects of using this method:

• It works by altering the constant from the homogeneous solution, allowing it to depend on \( x \).
• This creates an adaptable solution that fits the full equation.

The flexibility of this approach is what allows it to address parts of the equation that standard homogeneous solutions can't solve alone, thus finding the particular solution for complex differential equations.

Integrating Factor Method

The integrating factor method is a powerful tool used to transform linear differential equations into an easily solvable form.
By multiplying the entire differential equation by a strategically chosen function (integrating factor), it becomes possible to simplify the integration of the entire expression.
For an equation like \( y' - (1/2x)y = (1/2)x^2 - (1/2x) \), the integrating factor is \( e^{-ln|x|} \).
Major points of interest include:

• Multiplying by the integrating factor reconfigures the equation, making the left side a derivative of a product.
• It facilitates integration by creating a more straightforward path to find solutions.

Thus, with the integrating factor, the problem simplifies into an easily handled equation, leading directly to the desired specific solutions for our differential equation.