# (Cauchy-Euler Equation): Find the solution of the differential equation x^2y'' - 2xy' + 2y = 0 .

## Question:

(Cauchy-Euler Equation): Find the solution of the differential equation {eq}x^2y'' - 2xy' + 2y = 0 .{/eq}

## Cauchy-Euler Equation:

A Cauchy-Euler Equation is a differential equation of the form

{eq}a_{n}x^{n}y^{n}+a_{n-1}x^{n-1}y^{n-1}+....a_{1}x\acute{y}+a_{0}y=g\left ( x \right ) {/eq}

The homogeneous second order equation will be: {eq}ax^{2}{y}''+bx{y}'+cy=0 {/eq}

when we substitute {eq}y=x^{m}{/eq}

Then, the auxiliary equation is

{eq}am^{2}+\left ( b-a \right )m+c=0 {/eq}

There are 3 different cases to be considered, almost similar to

constant coefficients method:

Case I: Distinct Real Roots {eq}y=c_{1}x^{m_{1}}+c_{2}x^{m_{2}} {/eq}

Case II: Repeated Real Roots {eq}y=c_{1}x^{m_{1}}+c_{2}x^{m_{2}}lnx {/eq}

Case III: Conjugate Complex Roots {eq}y=x^{\alpha }\left [ c_{1}cos\left ( \beta\, lnx \right )+c_{2}sin\left ( \beta\,lnx \right ) \right ] {/eq}

## Answer and Explanation:

The differential equation {eq}x^2y'' - 2xy' + 2y = 0\\ a=1,b=-2,c=2 {/eq}

Write in auxiliary equation {eq}ax^{2}{y}''+bx{y}'+cy=0 \\ \Rightarrow m^{2}+\left ( -2-1 \right )m+2=0\\ \Rightarrow m^{2}-3m+2=0\\ \Rightarrow \left ( m-1 \right )\left ( m-2 \right )=0\\ \Rightarrow m=1,m=2 {/eq}

For Case I, the general solution is {eq}y=c_{1}x^{m_{1}}+c_{2}x^{m_{2}}\\ \Rightarrow y=c_{1}x+c_{2}x^{2} {/eq}

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