Question

Prove: If \(a, b, c, \alpha,\) and \(M\) are constants and \(M \neq 0\) then $$ a x^{2} y^{\prime \prime}+b x y^{\prime}+c y=M x^{\alpha} $$ has a particular solution \(y_{p}=A x^{\alpha}(A=\) constant \()\) if and only if \(a \alpha(\alpha-1)+b \alpha+c \neq 0\). If \(a, b, c,\) and \(\alpha\) are constants, then $$ a\left(e^{\alpha x}\right)^{\prime \prime}+b\left(e^{\alpha x}\right)^{\prime}+c e^{\alpha x}=\left(a \alpha^{2}+b \alpha+c\right) e^{\alpha x} $$

Short answer

Short answer: To prove that the second-order linear differential equation \(axy^{2}y^{\prime \prime}+bxy^{\prime}+cy = Mx^{\alpha}\) has a particular solution of the form \(y_p = A x^{\alpha}\) if and only if the condition \(a\alpha(\alpha-1)+b\alpha+c \neq 0\) holds, we follow these steps: 1. Find the derivatives of the proposed solution, \(y_p'= \alpha A x^{\alpha - 1}\) and \(y_p'' = \alpha(\alpha - 1)A x^{\alpha - 2}\). 2. Substitute the proposed solution and its derivatives into the given equation. 3. Simplify the equation to obtain \(a\alpha(\alpha - 1)A x^{\alpha} + b \alpha Ax^{\alpha} + cAx^{\alpha} = Mx^{\alpha}\). 4. Factor out the common term \(Ax^{\alpha}\) and set the coefficients equal to the RHS, leading to the condition: \(a\alpha(\alpha - 1) + b\alpha + c \neq 0\). 5. Address the specific case of the function \(e^{\alpha x}\) by recognizing that it is a special instance of the general framework of the problem, and the equation holds as previously shown. The condition \(a\alpha(\alpha-1)+b\alpha+c \neq 0\) ensures the existence of a particular solution of the form \(y_p = A x^{\alpha}\) for the given differential equation.

Step-by-step solution

Identify the given differential equation and its derivatives

The given differential equation is $$ axy^{2} y^{\prime \prime}+bxy^{\prime}+cy=Mx^{\alpha} $$ We are given the proposed particular solution \(y_p = A x^{\alpha}\), where \(A\) is a constant. Let's find the first and second derivatives of this function with respect to \(x\): $$ y_p' = \alpha A x^{\alpha - 1} $$ and $$ y_p'' = \alpha(\alpha - 1)A x^{\alpha - 2}. $$

Substitute the proposed solution and its derivatives into the given equation

Now we will substitute the proposed solution, \(y_p\), and its derivatives, \(y_p'\) and \(y_p''\), into the given differential equation: $$ a x^2(\alpha(\alpha - 1)A x^{\alpha - 2}) + bx(\alpha A x^{\alpha - 1}) + c(A x^{\alpha}) = Mx^{\alpha} $$

Simplify the equation

Let's simplify the equation by multiplying the terms and combining like terms: $$ a\alpha(\alpha - 1)A x^{\alpha} + b \alpha Ax^{\alpha} + cAx^{\alpha} = Mx^{\alpha} $$

Factor out the common term and derive the condition

Factor out the term \(Ax^{\alpha}\) on the LHS of the equation and set the coefficients equal to the RHS: $$ (Ax^{\alpha})(a\alpha(\alpha - 1) + b\alpha + c) = Mx^{\alpha} $$ Since \(Ax^{\alpha}\) is not equal to zero, we can divide both sides of the equation by \(Ax^{\alpha}\) : $$ a\alpha(\alpha - 1) + b\alpha + c = M $$ Now, for this equation to have a non-trivial solution (i.e., a solution where \(A\) is not zero), then the following condition must hold: $$ a\alpha(\alpha - 1) + b\alpha + c \neq 0 $$

Addressing the function \(e^{\alpha x}\)

The last part of the exercise states that if \(a, b, c,\) and \(\alpha\) are constants, then $$ a\left(e^{\alpha x}\right)^{\prime \prime}+b\left(e^{\alpha x}\right)^{\prime}+c e^{\alpha x}=\left(a \alpha^{2}+b \alpha+c\right)e^{\alpha x}. $$ Note that this equation is just a specific case of the initial differential equation where \(M = a\alpha^2 + b\alpha + c\), and \(y = e^{\alpha x}\). By substituting \(y = e^{\alpha x}\) (which is equivalent to \(y = x^\alpha\) with \(A = 1\)) into the given differential equation and simplifying, we can see that the given equation in this part of the exercise holds. Since we have already proven the main part of the exercise, we have also shown how this specific case fits into the general framework of the problem.

Key concepts

Particular Solution

In the realm of differential equations, a particular solution is a specific solution that satisfies a differential equation under given initial or boundary conditions. Unlike the general solution, which contains arbitrary constants and represents an entire family of solutions, a particular solution is a single, specific instance that fits the particular circumstances.
In the equation quoted: \[ a x^{2} y^{\'\prime\prime} + b x y^{\prime} + c y = M x^{\alpha}, \]proving that \(y_p = A x^{\alpha}\) is a particular solution involves ensuring it satisfies the equation when specific constants are provided. By substituting the proposed form into the differential equation and calculating its derivatives, we derive the condition under which this form satisfies the equation, which is:
\[ a\alpha(\alpha-1) + b\alpha + c = M. \]This particular expression provides us with a solution under the particular case that the coefficients satisfy non-zero conditions.

• Particular solutions are specific and unique to the initial conditions set by the differential equation.
• They illustrate how a single equation can be tailored to meet specific requirements.

Second Order Differential Equation

A second-order differential equation is one that involves the second derivative of a function. It takes the form: \[ a y^{\''} + by^{\'} + cy = h(x), \]where \(a\), \(b\), and \(c\) are coefficients, and \(h(x)\) is a function of \(x\). These types of equations are essential in modeling dynamics in physics, engineering, and many other scientific disciplines.
They characterize systems where the rate of change itself changes, adding a layer of complexity over first-order equations.
For the example exercise, we handle the equation: \[ a x^{2} y^{\prime\prime} + b x y^{\prime} + c y = M x^{\alpha}. \]
Second-order equations are sometimes more challenging to solve than first-order ones. They require careful considerations of both homogenous (simplified) and inhomogeneous (with added function) parts to find comprehensive solutions.

• They often describe oscillating systems, waves, and other natural phenomena.
• Solutions typically include both homogeneous and particular parts.

Homogeneous Equations

Homogeneous differential equations are a special category where the function equals zero throughout every variable relation, taking the form: \[ a y^{\''} + b y^{\'} + c y = 0. \]
Homogeneous equations lack external forces or inputs, meaning they model systems in equilibrium or internal dynamics in isolation.
Solving them involves finding complementary solutions that may be paired with particular solutions for non-homogeneous equations.
In the context of the exercise, when no external factor \(M x^{\alpha}\) is applied, the given differential equation becomes homogeneous. The solutions might involve exponential or polynomial functions that describe the natural behavior of the system.

• These equations serve as a step to understanding the comprehensive solution involving particular cases.
• They depict inherent dynamics devoid of external interference or constant term forces.

Exponential Function

The exponential function, often arising in solutions of differential equations, is defined as \(e^{x}\). It represents the continuous growth process and occurs frequently in physics, finance, and natural sciences.
Exponential functions exhibit constant rates of growth, described as self-referential, where the function's rate of increase corresponds to its value. In differential equations, they offer natural solutions to linear cases.
When solving with differentials, say: \[ a(e^{\alpha x})^{\prime\prime} + b(e^{\alpha x})^{\prime} + c e^{\alpha x} = (a \alpha^{2} + b \alpha + c)e^{\alpha x}, \]using the exponential function ensures rapid solution formulation due to its derivative properties: - Derivative of \(e^{\alpha x}\) is \(\alpha e^{\alpha x}\).
This property makes \(e^{x}\) a favorable candidate when modeling natural phenomena like decay or growth, fitting various parameters seamlessly.

• Exponential functions provide a reliable tool for solving differential equations.
• They integrate efficiently because of their straightforward derivative expressions.