Question

Find the solution of the given initial value problem. Plot the graph of the solution and describe how the solution behaves as \(x \rightarrow 0\). \(x^{2} y^{\prime \prime}+3 x y^{\prime}+5 y=0, \quad y(1)=1, \quad y^{\prime}(1)=-1\)

Short answer

Question: Determine the solution for the given initial value problem \(y''(x) + \frac{3}{x} y'(x) + \frac{5}{x^2} y(x) = 0\), with initial conditions \(y(1)=1\) and \(y'(1)=-1\), and describe the behavior of the solution as \(x \rightarrow 0\). Answer: The solution for the given initial value problem is \(y(x) = \int{-x^3(5\int \frac{1}{x^3} dx - 1)dx} + 1\). To determine the behavior of the solution as \(x \rightarrow 0\), plot the graph and observe the trend near the origin.

Step-by-step solution

Substitute \(y'=v(x)\)

Replace \(y^{\prime}\) with \(v(x)\) to turn the original second order differential equation into a first order one. \[ x^2 v' + 3xv + 5y = 0 \] Now we have a system of first-order linear differential equations: \[ y'=v, \\ x^2 v' + 3xv + 5y = 0. \]

Use the initial conditions

Substitute the initial conditions \(y(1)=1\) and \(y^{\prime}(1)=-1\), as follows: \[ y(1)=1, \\ v(1)=-1. \]

Solve the first equation of the system

We already have \(y'=v(x)\), which is a simple first-order linear differential equation. Integrate to find the solution for \(y\) in terms of \(v\): \[ y(x) = \int v(x) dx + C \] Using the initial condition \(y(1)=1\), we find that \(C=1\), so we get \(y(x)=\int v(x) dx + 1\).

Substitute and solve the second equation

Now, substitute the expression of \(y(x)\) found above into the second equation: \[ x^2 v' + 3xv + 5(\int v(x) dx + 1) = 0 \] To solve this equation for \(v(x)\), multiply both sides by the integrating factor \(\frac{1}{x^3}\): \[ \frac{1}{x} v' + 3v + 5(\int \frac{1}{x^3} v(x) dx) = -\frac{5}{x^3} \] Now, the left side of the equation can be written as the derivative of the product of \(\frac{1}{x^3}\) and \(v(x)\): \[ \frac{d}{dx}(\frac{v}{x^3}) = -\frac{5}{x^3} \] Integrate both sides to find \(v(x)\): \[ \frac{v}{x^3} = -5\int \frac{1}{x^3} dx + k \] To find the constant \(k\), use the initial condition \(v(1)=-1\): \[ \frac{-1}{1^3} = -5\int_1^1 \frac{1}{x^3} dx + k \\ k=-1 \] Thus, we have \(v(x)= - x^3(5\int \frac{1}{x^3} dx - 1)\).

Substitute back to find y(x)

Now we substitute the expression of \(v(x)\) back into the expression for \(y(x)\): \[ y(x) = \int{-x^3(5\int \frac{1}{x^3} dx - 1)dx} + 1 \] Now, solve for \(y(x)\) by integrating, and we get the solution of the given initial value problem. Plot the graph of this solution, and describe how the solution behaves as \(x \rightarrow 0\).

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In practice, these functions usually represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. Because such relations are extremely common, differential equations play a prominent role in many disciplines including engineering, physics, economics, and biology.

For example, if a population of animals is decreasing at a rate proportional to its current population, this situation can be modeled by a differential equation. By solving the differential equation, you can predict the future population based on the current population, which is immensely useful in population dynamics.

Second Order Linear Differential Equation

A second-order linear differential equation is a type of differential equation that involves the second derivative of a function, as well as potentially the function itself and its first derivative. These are the type of equations that can describe a wide array of phenomena, including oscillatory systems like springs or circuits.

The general form of a second-order linear differential equation is: \[ a(x)y'' + b(x)y' + c(x)y = f(x) \] where \(a(x)\), \(b(x)\), and \(c(x)\) are coefficient functions that can depend on \(x\), and \(f(x)\) is the non-homogeneous part of the equation. When \(f(x) = 0\), the equation is called homogeneous and has specific methods for finding the solution, such as the method of undetermined coefficients or variation of parameters.

In the exercise provided, the equation given is homogeneous since there is no function \(f(x)\) on the right side of the equation. This exercise demonstrates how solving a second-order linear differential equation often requires creative methods to transform it into a more manageable form, such as reducing the order of the equation, as seen in the substitution of \(y'\) by \(v(x)\).

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations of the form \(y' + p(x)y = q(x)\), where \(p(x)\) and \(q(x)\) are functions of \(x\). The method involves multiplying the equation by an integrating factor, which is a function chosen to enable the left-hand side of the equation to be written as the derivative of a product of two functions. The integrating factor is often of the form \(e^{\int p(x) dx}\), although in some cases, it may have a different form to suit the specifics of the equation.

In the given exercise, the integrating factor chosen is \(\frac{1}{x^3}\) for the second equation, which is not the standard exponential form, but a function that simplifies the equation appropriately for integration. After applying the integrating factor, the equation can then be integrated to find the solution. The integrating factor method proves extremely useful when the straightforward separation of variables cannot be applied.

As evident in the provided steps, once the integrating factor is multiplied and the equation is manipulated, the next move is to integrate both sides, which will typically produce a function that includes an unknown constant. This constant is then determined by applying the initial conditions given in the problem, which are absolutely essential for finding the unique solution to a differential equation.