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How to Find a Second Linearly Independent Solution for a Repeated Root in a Cauchy-Euler Equation

Have you ever come across a second-order differential equation that looks like this?

x²y″ − 3xy′ + 4y = 0

This is what's called a Cauchy-Euler equation (also known as an equidimensional equation). These are special because they match the pattern:

x²y″ + axy′ + by = 0

and we can solve them with a neat trick—try solutions of the form y = xʳ. But what happens when that method gives us a repeated root? Let's walk through it together!

Step 1: Assume a Power Function

Suppose we guess a solution:

y = xʳ

Then its derivatives are:

  • y′ = r·xr−1
  • y″ = r(r−1)·xr−2

Plug these into the equation:

x²y″ − 3xy′ + 4y = x²·r(r−1)xr−2 − 3x·r xr−1 + 4 xr

Simplify:

xʳ [r(r−1) − 3r + 4] = 0

This gives the characteristic equation:

r² − 4r + 4 = 0 → (r − 2)² = 0

We have a repeated root! r = 2

Wait... Isn't That Only One Solution?

Yes! The solution we get is:

y₁(x) = x²

But second-order differential equations need two linearly independent solutions to form the general solution.

When there's a repeated root, the second solution includes a logarithmic term:

y₂(x) = x² ln(x)

Where did that come from? Let’s find out.

Step 2: Use Reduction of Order

Suppose the second solution looks like:

y₂(x) = v(x) · x²

Differentiate:

  • y₂′ = v′·x² + 2v·x
  • y₂″ = v″·x² + 4v′·x + 2v

Now plug into the original differential equation:

x²y₂″ − 3x y₂′ + 4y₂

After substitution and simplification, the equation becomes:

v″·x⁴ + v′·x³ = 0

Divide by x³ (assuming x ≠ 0):

v″·x + v′ = 0

Step 3: Solve the Simpler Equation

This is now a first-order linear ODE in v′:

Let u = v′ ⇒ x du/dx + u = 0

Separate variables and integrate:

∫du/u = −∫dx/x ⇒ ln|u| = −ln|x| + C ⇒ u = C/x

Then:

v′ = C/x ⇒ v = C ln x + D

Step 4: Get the Final Answer

Now plug back:

y₂(x) = v(x) · x² = (C ln x + D) x²

So the two solutions are:

  • y₁(x) = x²
  • y₂(x) = x² ln x

The general solution is:

y(x) = C₁ x² + C₂ x² ln x

Final Thoughts

The Cauchy-Euler equation is one of those beautiful intersections of algebra and calculus. It teaches us that sometimes, when two solutions look the same, we need to dig deeper—literally into a logarithm—to find a new direction.

Now you’re ready to handle repeated roots like a pro.

Review Summary

  • Try y = xʳ — works when roots are real and distinct.
  • If you get a repeated root, the second solution involves a logarithmic term.
  • Use reduction of order to find it: set y₂ = v(x) xʳ and solve.
  • Final form: y₂ = xʳ ln x

Happy mathing!

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