Question

Finding a Particular Solution In Exercises \(17-24,\) find the particular solution of the differential equation that satisfies the initial condition. $$ \text{Differential Equation} \quad \text{Initial Condition} $$ $$ x^{3} y^{\prime}+2 y=e^{1 / x^{2}} \quad y(1)=e $$

Short answer

The particular solution of the given differential equation \(x^{3} y^{\prime}+2 y=e^{1 / x^{2}}\) satisfying the initial condition \(y(1)=e\) is \(y = \frac{1}{x^2}\int x^2 e^{1 / x^{2}} dx\), where the right side integral is evaluated so that it equals \(e\) at \(x=1\).

Step-by-step solution

Recognize the type of differential equation

The differential equation can be written in the standard form as: \(y'+P(x) y= Q(x)\) where \(P(x) = \frac{-2}{x^3}\) and \(Q(x) = \frac{e^{1 / x^{2}}}{x^3}\) . It is thus a first order linear non-homogeneous differential equation.

Find the integrating factor

The integrating factor is calculated as \(e^{\int P(x) dx}\). Substituting \(P(x) = \frac{-2}{x^3}\) and integrating, one finds that the integrating factor is \(\frac{1}{x^2}\).

Multiply through by the integrating factor

By multiplying each term by the integrating factor, the new differential equation becomes: \(x y'+2 y=x^2 e^{1 / x^{2}}\) which is now in the form \(\frac{d}{dx}[x^2 y] = x^2 e^{1 / x^{2}}\).

Integrate both sides

Integrating both sides of the equation gives: \(x^2 y = \int x^2 e^{1 / x^{2}} dx\). The integral on the right is non-trivial and cannot be expressed in terms of standard mathematical functions, and thus will be left as an indefinite integral.

Solve for y

After integrating, rearranging and solving the expression for \(y\), one gets: \(y = \frac{1}{x^2}\int x^2 e^{1 / x^{2}} dx\)

Apply the initial condition

Insert the initial condition \(y(1)=e\) into the solution above. Since the integral on the right-hand side cannot be solved explicitly, one can only say that the integral equals a certain value.

Key concepts

First Order Linear Differential Equations

First Order Linear Differential Equations form the core of many real-world problems. These equations are characterized by their first derivative, and can often be written in a standard form like this: \[ y' + P(x) y = Q(x) \]

• y' is the derivative of y with respect to x.
• P(x) is a function of x.
• Q(x) is another function of x, usually representing some external influence on the system.

These equations are called linear because they only involve linear terms, like y and y', and don't feature nonlinear expressions like y² or x·y. Recognizing this form allows us to apply systematic methods to find solutions. The simplicity and generality of first-order linear differential equations make them the entry point for learning about more complex differential equations.

Integrating Factor

An integrating factor is a clever mathematical tool that simplifies the process of solving a linear differential equation. The main idea is to multiply the entire equation by a specially chosen function that allows the left-hand side to be simplified into a single derivative. Here's how it works:

• Given our standard form, \(y' + P(x) y = Q(x)\), we calculate the integrating factor as \(e^{\int P(x) dx}\).
• Multiply through the entire equation by this integrating factor. This simplifies the left-hand side into the derivative of the product of the integrating factor and y, i.e., \(\frac{d}{dx} [\text{Integrating Factor} \, y]\).

This process takes advantage of product rule differentiation in reverse. With the equation now simpler, integrating both sides becomes straightforward. Although this strategy might appear abstract, it essentially transforms the differential equation into one that is easier to solve.

Initial Conditions

Initial conditions provide specific information that allows for finding particular solutions to differential equations. In general, a first-order differential equation's solution involves an arbitrary constant.

• To identify a precise solution, a given initial condition specifies the value of the function y at a certain point x.
• For example, with an initial condition like \(y(1) = e\), we know exactly what value y takes when \(x = 1\).

Without an initial condition, you'd only get a family of solutions. Applying initial conditions anchors the solution to a specific curve among these possibilities. It’s akin to aligning the solution with the exact scenario in a real-world problem. Using initial conditions is crucial for problems in physics and engineering where precise solutions are essential.

Non-homogeneous Equations

Non-homogeneous equations are differential equations in which the terms do not all depend on the function or its derivatives alone. These equations include an added term that can be seen as a source or external force, represented here by \(Q(x)\). In our given equation, \(x^3 y' + 2y = e^{1/x^2}\), the right-hand side \(e^{1/x^2}\) is the non-homogeneous part.

• The presence of this term indicates that the system or process it describes is being driven by an external factor.
• Solving these equations requires finding both the complementary (homogeneous) solution and a particular solution.

The added challenge in solving non-homogeneous equations is to appropriately account for this non-homogeneous term. Mastering this concept allows one to handle more dynamic systems in mathematical modeling and applied sciences.