Differential Equation Analysis

The given differential equation is:

\(x^2 \frac {d^2y} {dx^2} - 2x \frac {dy} {dx} + 2y = 0\)

This is a second-order linear homogeneous differential equation with variable coefficients. This specific type of equation is known as a Cauchy-Euler differential equation (also sometimes called an Euler-Cauchy equation). It has the general form \(ax^2 \frac {d^2y} {dx^2} + bx \frac {dy} {dx} + cy = 0\), where 'a', 'b', and 'c' are constants.

Cauchy-Euler Method Explained

To find the solution of a Cauchy-Euler differential equation, we use a specific method involving a substitution. The process typically involves the following steps:

1. Assume a solution of the form \(y = x^m\), where 'm' is a constant that we need to determine.
2. Calculate the first derivative of the assumed solution: \( \frac {dy} {dx} = mx^{m-1} \).
3. Calculate the second derivative of the assumed solution: \( \frac {d^2y} {dx^2} = m(m-1)x^{m-2} \).
4. Substitute these derivatives and \(y = x^m\) back into the original differential equation.
5. Simplify the resulting equation by factoring out \(x^m\). This will lead to an algebraic equation in terms of 'm', which is called the characteristic equation (or auxiliary equation).
6. Solve the characteristic equation for 'm'. The nature of these roots (real and distinct, real and repeated, or complex conjugate) dictates the specific form of the general solution to the differential equation.

Step-by-Step Solution

Let's apply the Cauchy-Euler method to solve the given differential equation:

\(x^2 \frac {d^2y} {dx^2} - 2x \frac {dy} {dx} + 2y = 0\)

1. Assume a Solution:

   We start by assuming a solution of the form:

   \(y = x^m\)

2. Calculate Derivatives:

   Next, we find the first and second derivatives of \(y = x^m\):

   The first derivative is:

   \( \frac {dy} {dx} = \frac{d}{dx}(x^m) = mx^{m-1} \)

   The second derivative is:

   \( \frac {d^2y} {dx^2} = \frac{d}{dx}(mx^{m-1}) = m(m-1)x^{m-2} \)

3. Substitute into the Differential Equation:

   Substitute \(y\), \(\frac{dy}{dx}\), and \(\frac{d^2y}{dx^2}\) back into the original differential equation:

   \(x^2 [m(m-1)x^{m-2}] - 2x [mx^{m-1}] + 2[x^m] = 0\)

   Simplify each term by combining the powers of \(x\):

   \(m(m-1)x^{2+m-2} - 2mx^{1+m-1} + 2x^m = 0\)

   \(m(m-1)x^m - 2mx^m + 2x^m = 0\)

4. Form the Characteristic Equation:

   Factor out the common term \(x^m\) from the entire equation. Since \(x^m \neq 0\) for a non-trivial solution, we can divide by \(x^m\):

   \(x^m [m(m-1) - 2m + 2] = 0\)

   This gives us the characteristic equation:

   \(m(m-1) - 2m + 2 = 0\)

   Expand and simplify the equation:

   \(m^2 - m - 2m + 2 = 0\)

   \(m^2 - 3m + 2 = 0\)

5. Solve the Characteristic Equation:

   We now solve this quadratic equation for 'm'. We can factor the quadratic expression:

   \((m-1)(m-2) = 0\)

   This yields two distinct real roots for 'm':

   \(m_1 = 1\)

   \(m_2 = 2\)

6. Write the General Solution:

   For a Cauchy-Euler differential equation with distinct real roots \(m_1\) and \(m_2\), the general solution is given by the formula:

   \(y = c_1x^{m_1} + c_2x^{m_2}\)

   Substitute the values of \(m_1 = 1\) and \(m_2 = 2\) into this formula:

   \(y = c_1x^1 + c_2x^2\)

   \(y = c_1x + c_2x^2\)

   Here, \(c_1\) and \(c_2\) are arbitrary constants determined by initial or boundary conditions (if provided).

Solution Comparison

Let's compare our derived general solution with the given options to find the correct match:

Our calculated solution, \(y = c_1x + c_2x^2\), perfectly matches Option 1.