Problem 2 - Advanced Engineering Mathematics

Question

Solve the given differential equation. $$ 4 x^{2} y^{\prime \prime}+y=0 $$

Step-by-Step Solution

Answer

The general solution is \(y(x) = C_1x^{1/2} + C_2x^{1/2}\ln(x)\).

Step 1: Identify the Type of Differential Equation

The given differential equation is \(4x^2y'' + y = 0\). This is a second-order linear homogeneous differential equation with variable coefficients.

Step 2: Apply the Method for Second-Order Homogeneous Equations

For a differential equation of the form \(x^2y'' + py' + qy = 0\), where \(p(x)\) and \(q(x)\) can depend on \(x\), we assume a solution of the form \(y = x^m\). This method is known as the Cauchy-Euler method or equidimensional equations method.

Step 3: Substitute the Trial Solution

Substitute \(y = x^m\) into the differential equation. The derivatives are \(y' = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\). Substituting these into the original equation gives:\[4x^2(m(m-1)x^{m-2}) + x^m = 0\]

Step 4: Simplify the Equation

Simplify the expression from the previous step:\[4m(m-1)x^m + x^m = 0\] Factor out \(x^m\):\[x^m(4m(m-1) + 1) = 0\] Since \(x^m eq 0\), the factor must equal zero: \[4m(m-1) + 1 = 0\]

Step 5: Solve the Characteristic Equation

Solve the equation \(4m(m-1) + 1 = 0\):\[4m^2 - 4m + 1 = 0\] This is a quadratic equation. Use the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -4\), and \(c = 1\):\[m = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}\]\[m = \frac{4 \pm \sqrt{16 - 16}}{8}\]\[m = \frac{4}{8} = \frac{1}{2}\]

Step 6: Write the General Solution

Since the characteristic equation has a double root \(m = \frac{1}{2}\), the general solution for the differential equation is:\[y(x) = C_1x^{1/2} + C_2x^{1/2}\ln(x)\]Where \(C_1\) and \(C_2\) are constants.

Key Concepts

Second-order Differential Equation

A second-order differential equation involves the second derivative of a function, often denoted as \(y''\) in relation to its independent variable. In many practical problems, such equations describe how systems behave when acceleration (which is the second derivative of position) is involved.
In the given exercise, you see the equation \(4x^2y'' + y = 0\), where \(y''\) represents the second derivative of \(y\) concerning \(x\). This kind of equation introduces complexity beyond what first-order equations provide, as the solution approach significantly differs. The focus here is finding functions that satisfy the entire equation, involving analyzing both the nature and the implications of these derivatives.
Second-order differential equations can belong to a variety of types, but in this exercise, it's essential to note its linearity and the presence of variable coefficients, which lead it to a specialized solving technique known as the Cauchy-Euler method. This gives us a powerful toolkit for handling such equations.

Homogeneous Differential Equation

A differential equation is termed 'homogeneous' if all terms are a function of the unknown function and its derivatives, without any standalone constant terms. In other words, for the differential equation \(4x^2y'' + y = 0\), the expressed terms are solely combinations of \(y\), its derivatives, and the independent variable \(x\), all equating to zero.
The importance of identifying a differential equation as homogeneous lies in the solving strategy. Homogeneous equations, especially linear ones, can be treated with specific methods due to their forms, allowing for solutions that combine various standard functions.
In this problem, recognizing the equation as both homogeneous and linear allows us to employ strategies like assuming solutions of a specific form, such as \(y = x^m\), which simplifies the problem into a task of determining these parameters, rather than directly deriving \(y\) through integration as seen with non-homogeneous forms.

Variable Coefficients

Differential equations can have constant or variable coefficients. In our equation, \(4x^2y'' + y = 0\), the coefficient of the second derivative \(y''\) is \(4x^2\), which is dependent on \(x\). This makes them variable coefficients.
Variable coefficients significantly alter how differential equations are tackled because they require more nuanced solving methods. They don't comply with simpler techniques used for constant coefficients, like the typical constant coefficient linear differential equation methods.
In solving Cauchy-Euler differential equations, we inherently deal with variable coefficients. Here, the method involves transforming the equation into a form that can be solved as if it were a constant coefficient equation by assuming a trial solution of the form \(y = x^m\). This exploits the properties of exponents, reducing the problem complexity by factoring in the changing influence of \(x\) within the coefficients.

Characteristic Equation

The characteristic equation is a vital concept when dealing with differential equations, especially in the context of linear homogeneous equations like our example. It originates from the Cauchy-Euler method, where one assumes a solution of the form \(y = x^m\), substituting this into the original equation results in an algebraic equation in terms of \(m\).
In this exercise, we simplified the differential equation by substituting derivatives based on \(y = x^m\) which led us to \(x^m(4m(m-1) + 1) = 0\). Since \(x^m eq 0\), we focused on solving \(4m(m-1) + 1 = 0\) for \(m\).
The result is a quadratic equation known as the characteristic equation \(4m^2 - 4m + 1 = 0\), which is solved using the quadratic formula. For our equation, obtaining the value of \(m\) gives us necessary roots used to construct the general solution of the original differential equation.