Question

$$ y^{\prime \prime}+x y^{\prime}+y=0 $$

Short answer

The given second-order non-linear ODE: \(y^{\prime\prime} + xy^{\prime} + y = 0\), can be converted into a system of first-order linear ODEs by introducing a new variable \(u(x) = y^{\prime}(x)\): \(\begin{cases} y^{\prime}(x) = u(x) \\ u^{\prime}(x) = -x u(x) - y(x) \end{cases}\) By solving this system, we get a general solution for y(x): \(y(x) = \int e^{-\left(\frac{1}{2}x^2\right)} \left( -\int y(x) e^{\left(\frac{1}{2}x^2\right)} dx + C \right) dx\) The specific solution will depend on the additional conditions provided in the problem, such as initial values or boundary conditions.

Step-by-step solution

Introduce a new variable u(x)

Let's introduce a new variable u(x) such that: $$ u(x) = y^{\prime}(x) $$ Now, we can rewrite the given equation as: $$ y^{\prime\prime}(x) + x u(x) + y(x) = 0 $$

Rewrite the equation as a system of first-order linear ODEs

Now, let's rewrite the equation as a system of first-order linear ODEs for the variables u(x) and y(x): $$ \begin{cases} y^{\prime}(x) = u(x) \\ u^{\prime}(x) = -x u(x) - y(x) \end{cases} $$

Solve the first ODE

We can solve the first ODE directly: $$ y^{\prime}(x) = u(x) $$ We can integrate it to get y(x): $$ y(x) = \int u(x) dx $$

Solve the second ODE

Now, we have to solve the second ODE: $$ u^{\prime}(x) + x u(x) = -y(x) $$ This is a first-order linear ODE in u(x). We can find an integrating factor: $$ I(x) = e^{\left(\int x dx\right)} = e^{\left(\frac{1}{2}x^2\right)} $$ Multiply both sides of the ODE by the integrating factor: $$ e^{\left(\frac{1}{2}x^2\right)} u^{\prime}(x) + x e^{\left(\frac{1}{2}x^2\right)} u(x) = -y(x)e^{\left(\frac{1}{2}x^2\right)} $$ Now, the left side of the equation becomes the derivative of the product: $$ \frac{d}{dx} \left( u(x)e^{\left(\frac{1}{2}x^2\right)} \right) = -y(x)e^{\left(\frac{1}{2}x^2\right)} $$ Integrate both sides with respect to x: $$ u(x) e^{\left(\frac{1}{2}x^2\right)} = -\int y(x) e^{\left(\frac{1}{2}x^2\right)} dx + C $$ Solve u(x): $$ u(x) = e^{-\left(\frac{1}{2}x^2\right)} \left( -\int y(x) e^{\left(\frac{1}{2}x^2\right)} dx + C \right) $$

Express y(x)

Recall that y(x) can be expressed as: $$ y(x) = \int u(x) dx $$ Now, use the expression for u(x) found in Step 4: $$ y(x) = \int e^{-\left(\frac{1}{2}x^2\right)} \left( -\int y(x) e^{\left(\frac{1}{2}x^2\right)} dx + C \right) dx $$ This is a complicated integral that might not have a closed-form solution. The solution, y(x), will depend on the specific conditions of the problem, such as initial values or boundary conditions.

Key concepts

Integrating Factor Method

The integrating factor method is a powerful technique used to solve first-order linear ordinary differential equations (ODEs). When faced with a differential equation in the form of \( y' + p(x)y = q(x) \) where \( p(x) \) and \( q(x) \) are functions of \( x \), finding an appropriate integrating factor can simplify the process significantly.

An integrating factor, denoted as \( I(x) \) is usually of the form \( e^{\int p(x)dx} \) and, when multiplied to both sides of the ODE, it allows us to write the left side as a product derivative of \( y \) and \( I(x) \) hence enabling the integration of both sides easily. This method encapsulates the magic of transforming a difficult-to-solve equation into a more tractable form, by essentially un-doing the differentiation that has happened and then integrating to find the original function.

System of First-Order ODEs

Transforming a higher-order ODE into a system of first-order ODEs is a strategic approach often employed to simplify complex problems. Each first-order equation in the system represents a component of the original ODE, often allowing for a straightforward integration process.

The solution to a single second-order ODE can be understood as a trajectory in a two-dimensional space, dictated by its position and velocity. When it's rewritten as a system of first-order ODEs, we tackle each dimension independently, effectively reducing the problem to a series of smaller, solvable puzzles. This divide-and-conquer strategy not only simplifies calculations but also opens the door to powerful analytical tools, such as phase plane analysis, and is key to applying numerical methods for more complicated ODEs where analytical solutions are hard to find.

Linear Differential Equations

Linear differential equations form a vast and fundamental category within the broader class of ODEs. These equations are characterized by the property that the dependent variable and its derivatives appear linearly - meaning terms involving the dependent variable are not multiplied or composed with each other.

The beauty of linear equations lies in their predictability: their solutions can be superimposed, and their behavior is easier to generalize. This relative simplicity and the linear superposition principle allow us to construct general solutions as combinations of particular solutions and homogeneous solutions. Importantly, tackling linear ODEs often opens the way to understanding more complex, nonlinear systems by providing insights into their behavior near equilibrium points.

Ordinary Differential Equations

At the heart of many scientific and engineering problems are ordinary differential equations (ODEs). These equations relate functions to their derivatives, providing a description of how a system changes with respect to one independent variable, typically time or space.

ODEs are categorized by their order, the highest derivative present, and can be further classified as linear or nonlinear, homogenous or nonhomogenous, and so on. Solving an ODE involves finding a function or set of functions that satisfy the equation for all values in a certain domain. This solution process requires the understanding of various mathematical concepts, including integration, and in the case of more complex or higher-order ODEs, the ability to simplify the problem into more manageable parts. ODEs are foundational not only in mathematics but form the building blocks for models in physics, biology, economics, and beyond.