# Show that y_{1} = x^{2} and y_{2} = x^{3} are solutions to the equation below that both satisfy...

## Question:

Show that {eq}y_{1} = x^{2} {/eq} and {eq}y_{2} = x^{3} {/eq} are solutions to the equation below that both satisfy {eq}y(0) = 0 {/eq} and {eq}y'(0) = 0 {/eq}.

$$x^{2}y'' - 4xy' + 6y = 0.$$

## Second-Order Cauchy-Euler Equations:

Suppose that {eq}a {/eq}, {eq}b {/eq}, and {eq}c {/eq} are real constants with {eq}a\ne 0 {/eq}. Then the equation

{eq}ax^2y''+bxy'+cy=0 {/eq}

is called a second-order homogeneous Cauchy-Euler equation.

A Cauchy-Euler equation has a singular point at {eq}x=0 {/eq}, because the coefficient {eq}ax^2 {/eq} of the highest-order derivative {eq}y'' {/eq} vanishes when {eq}x=0 {/eq}. So the standard theorems about existence and uniqueness of solutions to initial value problems do not apply to Cauchy-Euler equations given initial data at {eq}x=0 {/eq}.

## Answer and Explanation:

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Differentiating the function {eq}y_1(x)=x^2 {/eq}, we have:

{eq}\begin{align*} y_1'&=2x\\ y_1''&=2 \, . \end{align*} {/eq}

Substituting these...

See full answer below.

#### Learn more about this topic: First-Order Linear Differential Equations

from

Chapter 16 / Lesson 3
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In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.