Question

Find the solution of the given initial value problem. Plot the graph of the solution and describe how the solution behaves as \(x \rightarrow 0\). \(x^{2} y^{\prime \prime}-3 x y^{\prime}+4 y=0, \quad y(-1)=2, \quad y^{\prime}(-1)=3\)

Short answer

The given problem involves a second-order linear homogeneous differential equation. 2) What is the specific solution of the given differential equation with the initial conditions? The specific solution is y(x) = x^2 - x^3. 3) Describe the behavior of the graph as x approaches 0. As x approaches 0, the function behaves like x^2 and approaches 0.

Step-by-step solution

Identify the form of the differential equation

The given differential equation is a Cauchy-Euler equation, which can be written as: \(x^{2} y'' - 3x y' + 4y = 0\) This is a second-order linear homogeneous differential equation. Now, we will proceed to solve this equation.

Guess the form of the solution

Let's make a guess for the form of the solution for this Cauchy-Euler equation: \(y(x) = x^{m}\), where \(m\) is an exponent to be determined.

Derive the first and second derivatives of the guess

We need to find the first and second derivatives of \(y(x)\), and plug them into the given differential equation: \(y'(x) = m x^{m-1}\) \(y''(x) = m(m-1) x^{m-2}\)

Substitute the expressions of \(y\), \(y'\), and \(y''\) into the given differential equation

Substituting into the given differential equation, we get: \(x^{2} (m(m-1) x^{m-2}) - 3x (m x^{m-1}) + 4 (x^m) = 0\) Simplifying the equation, we get: \(m(m-1) x^m - 3m x^m + 4 x^m = 0\) Now, we can factor out \(x^m\): \(x^m (m(m-1) - 3m + 4) = 0\)

Solve for \(m\)

Since \(x^m \neq 0\), we can set the expression inside the parenthesis equal to zero: \(m^2 - m - 3m + 4 = 0\) \(m^2 - 4m + 4 = 0\) This quadratic equation can be factored as \((m-2)^2 = 0\). Thus, we have a double root: \(m = 2\)

Write out the general solution

Since we have a double root, we can write the general solution as: \(y(x) = c_1 x^2 + c_2 x^3\), where \(c_1\) and \(c_2\) are constants.

Apply the initial conditions

Now we will apply the initial conditions given in the problem: 1) \(y(-1) = 2\): \(2 = c_1 (-1)^2 + c_2 (-1)^3 \Rightarrow 2 = c_1 - c_2\) 2) \(y'(-1) = 3\): \(3 = 2c_1 (-1) + 3c_2 (-1)^2 \Rightarrow 3 = -2c_1 + 3c_2\) Now, we have the system of linear equations: $ \begin{cases} c_1 - c_2 = 2 \\ -2c_1 + 3c_2 = 3 \end{cases} $ Solving this system, we find that \(c_1 =1\) and \(c_2 = -1\).

Write out the specific solution

Now that we have found the values for \(c_1\) and \(c_2\), we can write out the specific solution: \(y(x) = x^2 - x^3\)

Plot the graph of the solution and describe behavior as \(x \rightarrow 0\)

Now, we can plot the graph of this function using any mathematical plotting software or tool, such as Desmos or WolframAlpha. As \(x\) approaches 0, the dominant term in the solution is the \(x^2\) term, as the \(x^3\) term becomes smaller in magnitude. Therefore, as \(x \rightarrow 0\), the function behaves like \(x^2\) and approaches 0.

Key concepts

Initial Value Problem

An initial value problem in the context of differential equations is a type of problem where not only a differential equation is given, but also the values of the solution (and possibly its derivatives) at a certain point, known as 'initial conditions'. These conditions are crucial as they help in finding a specific solution to the differential equation. When given a second-order differential equation, like in our exercise, you generally have two initial conditions: one for the function and one for its first derivative.

To solve an initial value problem, one typically first tries to find the general solution to the differential equation, which contains arbitrary constants. Afterward, these constants are determined by applying the initial conditions. In your exercise, after finding the general solution, the initial conditions, which in this case are, \(y(-1)=2\) and \(y'(-1)=3\), are used to compute the values of \(c_1\) and \(c_2\).

Second-Order Linear Homogeneous Differential Equation

A second-order linear homogeneous differential equation is characterized by its highest derivative being of the second order and the equation itself not having any terms that are not product of a constant and an unknown function or its derivatives. In simpler terms, there are no 'free-standing' constants or functions that are not multiplied by the unknown function we are trying to find.

These equations follow the form \(a(x)y'' + b(x)y' + c(x)y = 0\), where \(y''\) and \(y'\) represent the second and first derivatives of \(y\), respectively. The Cauchy-Euler equation falls into this category but has a distinguishing feature: the coefficients are powers of \(x\). The solution to such equations typically invloves finding the roots of the characteristic equation, which leads to determining the form of the solution for \(y(x)\). When solving, we notice that if the characteristic equation has a repeated root, our general solution will include an extra term that accounts for this multiplicity, which is why in the exercise solution, for a double root \(m=2\), the term \(c_2x^3\) appears alongside \(c_1x^2\).

Characteristic Equation

The characteristic equation is a powerful tool when dealing with linear homogeneous differential equations, especially for those with constant coefficients. By transforming the differential equation into an algebraic one, we can analyze the roots which give us a direct way to write down the general solution of the differential equation.

In the context of the Cauchy-Euler equation, the characteristic equation is obtained by substituting \(y(x) = x^m\) into the differential equation and simplifying, which typically results in a polynomial where \(m\) is the variable. In the exercise, for example, the characteristic equation turned out to be \((m-2)^2 = 0\), which means \(m = 2\) is a double root. This repeated root impacts the structure of the general solution, as it would inform us to include a term that reflects the multiplicity of the root, resulting in a complete solution that adheres to the form required for a Cauchy-Euler equation.