Differential Equation Solution: Cauchy-Euler Method

The given differential equation is of the form \( \displaystyle {x^2}\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} + y = \log x \). This is a second-order linear non-homogeneous differential equation with variable coefficients, specifically a Cauchy-Euler (or Euler-Cauchy) equation.

To solve this type of equation, we typically use the substitution \(x = e^z\), which implies \(z = \log x\). This transforms the equation into a linear differential equation with constant coefficients, which is easier to solve.

Transformation of Derivatives

Let's find the transformed derivatives in terms of \(z\):

• For the first derivative: \[ \frac{{dy}}{{dx}} = \frac{{dy}}{{dz}} \cdot \frac{{dz}}{{dx}} = \frac{{dy}}{{dz}} \cdot \frac{1}{x} = \frac{1}{{{e^z}}}\frac{{dy}}{{dz}} = {e^{ - z}}\frac{{dy}}{{dz}} \] Therefore, \( \displaystyle x\frac{{dy}}{{dx}} = x\left( {\frac{1}{x}\frac{{dy}}{{dz}}} \right) = \frac{{dy}}{{dz}} \).
• For the second derivative: \[ \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{1}{x}\frac{{dy}}{{dz}}} \right) \] Applying the product rule: \[ = - \frac{1}{{{x^2}}}\frac{{dy}}{{dz}} + \frac{1}{x}\frac{d}{{dx}}\left( {\frac{{dy}}{{dz}}} \right) \] Now, apply the chain rule to \( \displaystyle \frac{d}{{dx}}\left( {\frac{{dy}}{{dz}}} \right) \): \[ \frac{d}{{dx}}\left( {\frac{{dy}}{{dz}}} \right) = \frac{d}{{dz}}\left( {\frac{{dy}}{{dz}}} \right) \cdot \frac{{dz}}{{dx}} = \frac{{{d^2}y}}{{d{z^2}}} \cdot \frac{1}{x} \] Substitute this back: \[ \frac{{{d^2}y}}{{d{x^2}}} = - \frac{1}{{{x^2}}}\frac{{dy}}{{dz}} + \frac{1}{x}\left( {\frac{1}{x}\frac{{{d^2}y}}{{d{z^2}}}} \right) = - \frac{1}{{{x^2}}}\frac{{dy}}{{dz}} + \frac{1}{{{x^2}}}\frac{{{d^2}y}}{{d{z^2}}} \] Therefore, \( \displaystyle {x^2}\frac{{{d^2}y}}{{d{x^2}}} = {x^2}\left( {\frac{1}{{{x^2}}}\frac{{{d^2}y}}{{d{z^2}}} - \frac{1}{{{x^2}}}\frac{{dy}}{{dz}}} \right) = \frac{{{d^2}y}}{{d{z^2}}} - \frac{{dy}}{{dz}} \).

Transformed Differential Equation

Substitute the transformed derivatives and \( \log x = z \) into the original equation:

\( \displaystyle \left( {\frac{{{d^2}y}}{{d{z^2}}} - \frac{{dy}}{{dz}}} \right) - \left( {\frac{{dy}}{{dz}}} \right) + y = z \)

Simplifying, we get a linear differential equation with constant coefficients:

\( \displaystyle \frac{{{d^2}y}}{{d{z^2}}} - 2\frac{{dy}}{{dz}} + y = z \)

Let's represent the derivatives with the operator \(D = \frac{d}{dz}\):

\( (D^2 - 2D + 1)y = z \)

Solving the Homogeneous Equation (Complementary Function)

The homogeneous part is \( (D^2 - 2D + 1)y = 0 \).

The auxiliary equation is \( m^2 - 2m + 1 = 0 \).

This factors as \( (m - 1)^2 = 0 \).

So, the roots are \( m = 1, 1 \) (repeated real roots).

The complementary function \( y_c \) is:

\( y_c = (C_1 + C_2 z)e^z \)

Where \(C_1\) and \(C_2\) are arbitrary constants.

Finding the Particular Integral

We need to find \( y_p \) for \( Q(z) = z \).

\( y_p = \frac{1}{{{D^2} - 2D + 1}}z = \frac{1}{{{{(D - 1)}^2}}}z \)

To evaluate this, we can rewrite the denominator as \( (1 - D)^2 \) and use the binomial expansion \({(1-X)^{-n}} = 1 + nX + \frac{n(n+1)}{2!}X^2 + \dots \):

\( y_p = \frac{1}{{{{(1 - D)}^2}}}z = {(1 - D)^{ - 2}}z \)

Expand \({(1 - D)^{ - 2}}\) up to terms that will not vanish when operating on \(z\). Since \(z\) is a first-degree polynomial, \(D^2(z)\) and higher derivatives will be zero.

\( {(1 - D)^{ - 2}} = 1 + 2D + \frac{(-2)(-2-1)}{2!}D^2 + \dots = 1 + 2D + 3D^2 + \dots \)

Apply this operator to \(z\):

\( y_p = (1 + 2D + 3D^2 + \dots)z \)

\( y_p = 1 \cdot z + 2 \cdot D(z) + 3 \cdot D^2(z) \)

\( y_p = z + 2 \cdot 1 + 3 \cdot 0 \)

\( y_p = z + 2 \)

General Solution in Terms of z

The general solution is the sum of the complementary function and the particular integral:

\( y = y_c + y_p \)

\( y = (C_1 + C_2 z)e^z + z + 2 \)

Converting Back to x

Substitute back \(z = \log x\) and \(e^z = x\):

\( y = (C_1 + C_2 \log x)x + \log x + 2 \)

This is the final solution for the given differential equation.

Matching with Options

Comparing our solution \( y = (C_1 + C_2 \log x)x + \log x + 2 \) with the given options, it perfectly matches Option 2:

\( \text{y = (C}_1 \text{ + C}_2 \text{ logx)x + logx + 2, where C}_1 \text{ and C}_2 \text{ are arbitrary constants.} \)