Question

Auxiliary Equation for the Cauchy Euler Equation \( (x^2 D^2 + xD + 4)y = 0 \) is ____ .

• A. \( \theta^2 + 4 = 0 \)
• B. \( \theta^2 + 9 = 0 \)
• C. \( \theta^2 + 16 = 0 \)
• D. \( \theta^2 + 25 = 0 \)

Hint:

Cauchy-Euler Auxiliary Equation. For \( ax^2 y'' + bxy' + cy = 0 \), substitute \(y=x^m\) to get the auxiliary equation in m: \( am(m-1) + bm + c = 0 \). Alternatively, use substitution \(x=e^t\), \(\theta=d/dt\), where \(xD \rightarrow \theta\) and \(x^2D^2 \rightarrow \theta(\theta-1)\), leading to an auxiliary equation in \(\theta\).

Answer 1

The Correct Option is A

The given Cauchy-Euler equation is \( (x^2 D^2 + xD + 4)y = 0 \), where \(D = d/dx\)
To find the auxiliary equation, we use the substitution \(x = e^t\) or \(t = \ln x\)
Let \(\theta = d/dt\)
The transformations are: \( xD = \theta \) \( x^2 D^2 = \theta(\theta - 1) \) Substitute these into the operator: $$ \theta(\theta - 1) + \theta + 4 $$ $$ \theta^2 - \theta + \theta + 4 $$ $$ \theta^2 + 4 $$ The auxiliary equation in terms of \(\theta\) is obtained by setting this operator equal to zero: $$ \theta^2 + 4 = 0 $$ Alternatively, if using the substitution \(y=x^m\), the auxiliary equation in \(m\) is obtained by replacing \(x^2D^2\) with \(m(m-1)\) and \(xD\) with \(m\): $$ m(m-1) + m + 4 = 0 $$ $$ m^2 - m + m + 4 = 0 $$ $$ m^2 + 4 = 0 $$ Both methods lead to an auxiliary equation of the form (Variable)\(^2 + 4 = 0\)
Option (1) matches this form using \(\theta\)