Question

For each of the following equations, one solution \(u\) is given. Find the other solution by assuming a solution of the form \(y=u v\). \(x^{2} y^{\prime \prime}-3 x y^{\prime}+4 y=0 ; u=x^{2}\)

Short answer

The other solution is \[ y_2 = x^2 \ln|x| \].

Step-by-step solution

Understand the given equation

The given differential equation is \[ x^2 \frac{d^2 y}{dx^2} - 3x \frac{dy}{dx} + 4y = 0 \]and one of the solutions is given as \[ u = x^2 \].The objective is to find the other solution.

Assume the solution form

Assume the solution in the form \[ y = uv \].Since one solution is \[ u = x^2 \],assume \[ y = x^2 v \].

Find the first and second derivatives

First, find the first derivative of the assumed solution \[ y = x^2 v \].Using the product rule:\[ y' = \frac{d}{dx}(x^2 v) = x^2 \frac{dv}{dx} + 2x v \].Next, find the second derivative using the derivative of the first derivative:\[ y'' = \frac{d}{dx}(x^2 v' + 2x v) = x^2 \frac{d^2 v}{dx^2} + 4x \frac{dv}{dx} + 2v \].

Substitute derivatives into the original equation

Substitute the derivatives back into the original differential equation \[ x^2 y'' - 3x y' + 4y = 0 \]:\[ x^2 (x^2 \frac{d^2 v}{dx^2} + 4x \frac{dv}{dx} + 2v) - 3x (x^2 \frac{dv}{dx} + 2x v) + 4(x^2 v) = 0 \].

Simplify the equation

Simplify the equation by distributing and combining like terms:\[ x^4 \frac{d^2 v}{dx^2} + 4x^3 \frac{dv}{dx} + 2x^2 v - 3x^3 \frac{dv}{dx} - 6x^2 v + 4x^2 v = 0 \],which simplifies to\[ x^4 \frac{d^2 v}{dx^2} + x^3 \frac{dv}{dx} = 0 \].

Solve the simplified equation

Divide through by \[ x^3 \](not zero) to simplify the equation further:\[ x \frac{d^2 v}{dx^2} + \frac{dv}{dx} = 0 \].This is a first-order linear differential equation. Let \[ \frac{dv}{dx} = w \], so\[ x \frac{dw}{dx} + w = 0 \]or\[ \frac{dw}{dx} = -\frac{w}{x} \].

Integrate to find the solution

Separate the variables and integrate:\[ \frac{dw}{w} = -\frac{dx}{x} \],which upon integrating gives\[ \ln|w| = -\ln|x| + C_1 \],or\[ w = \frac{C_2}{x} \],where \[ C_2 \] is a constant. Since \[ w = \frac{dv}{dx} \],we have\[ \frac{dv}{dx} = \frac{C_2}{x} \].Integrate again to find\[ v = C_2 \ln|x| + C_3 \],where \[ C_3 \] is a constant.

Form the general solution

Since \[ y = x^2 v \] and \[ v = C_2 \ln|x| + C_3 \],the general solution is:\[ y = x^2 (C_2 \ln|x| + C_3) \].

Verify and conclude the other solution

A specific solution when not considering the constant \[C \] can be derived as:\[ y_2 = x^2 \ln|x| \].

Key concepts

Second-Order Differential Equations

In mathematics, a second-order differential equation is an equation that relates a function with its first and second derivatives. These equations often represent physical phenomena such as motion, electrical circuits, or fluid dynamics. For example, the given equation \[x^2 \frac{d^2 y}{dx^2} - 3x \frac{dy}{dx} + 4y = 0 \] is a second-order differential equation because it involves the second derivative \(\frac{d^2 y}{dx^2}\). Understanding these types of equations is critical for solving various real-world problems that involve change rates and dependencies.

General Solutions

The general solution of a differential equation encompasses all possible solutions that satisfy the given equation. For second-order equations like our example, it typically involves two arbitrary constants. The provided solution already gives one part: \( u = x^2 \). To find the complete general solution, we solve for another function using specific techniques, which in this case involves assuming a solution form \( y = uv \). Through systematic manipulation and solving, we find the other part of the general solution: \( y = x^2(C_2 \ln|x| + C_3) \).

Integration Techniques

Integration is a fundamental tool for solving differential equations. In this example, once we simplified to \[x \frac{d^2 v}{dx^2} + \frac{dv}{dx} = 0\], we used substitution and separation of variables to solve. Starting with \( \frac{dv}{dx} = w \), we derived \( \frac{dw}{dx} = -\frac{w}{x} \). By separating variables and integrating, we obtained \[ \ln|w| = -\ln|x| + C_1 \], leading to \[ w = \frac{C_2}{x} \]. Further integration gave us \[ v = C_2 \ln|x| + C_3 \], which are essential steps to combine for our general solution: \(y = x^2 (C_2 \ln|x| + C_3)\).

Homogeneous Equations

A differential equation is called homogeneous if every term is a multiple of the dependent variable or its derivatives. Mathematically, it's of the form \(a(x)y'' + b(x)y' + c(x)y = 0\). In our example, \[x^2 y'' - 3x y' + 4y = 0\], each term is indeed proportional to the dependent variable \(y\) and its derivatives, making it homogeneous. Solutions for homogeneous equations often involve finding specific solution forms that satisfy the entire equation, leading us to solutions that are typically combinations of such forms.