# When the polynomial associated with a second-order Cauchy-Euler equation has the repeated root r...

## Question:

When the polynomial associated with a second-order Cauchy-Euler equation has the repeated root {eq}r = r_1{/eq}, the second linearly independent solution is {eq}y = t^{r_1} ln t{/eq}, for {eq}t > 0{/eq}. Find the general solution of the following equations.

52. {eq}t^2 y'' - ty' + y = 0 {/eq}

53. {eq}t^2 y'' + 3ty' + y = 0 {/eq}

54. {eq}t^2 y'' - 3ty' + 4y = 0 {/eq}

55. {eq}t^2 y'' + 7ty' + 9y = 0{/eq}

## Cauchy-Euler Equation:

A homogeneous Euler-Cauchy equation of second order is a linear ordinary differential equation with variable coefficients of the form {eq}ax^2y'' +bxy' +cy = 0. {/eq} Its auxiliary equation is {eq}am(m-1) +bm +c = 0. {/eq} The transformation {eq}x=e^t {/eq} transforms it into a DE with constant coefficients. We find the roots of the auxiliary equation to obtain two values of m, {eq}m_1, \ m_2 {/eq}. If the roots are real and repeated, the general solution is given as:

{eq}x^m(A+b \ln x), {/eq} where A and B are arbitrary constants.

## Answer and Explanation: 1

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View this answer{eq}t^2 y'' - ty' + y = 0 {/eq}

The corresponding auxiliary equation is

{eq}m^2-m-m+1=0\\ m^2-2m+1=0\\ (m-1)^2=0\\ m=1, \ 1 {/eq}

Hence, the...

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Chapter 13 / Lesson 6This lesson is an introduction to differential calculus, the branch of mathematics that is concerned with rates of change. If you ever wanted to know how things change over time, then this is the place to start!