Recipe

Solve a Cauchy-Euler equation

Use when: the equation has the form ax^2y''+bxy'+cy=0 — the coefficient of y'' is x^2 and the coefficient of y' is x, each times a constant, for x>0

Goal: the general solution y(x), built from powers of x rather than exponentials

Taught in The Cauchy-Euler equation · a later section extends this recipe with more cases

1. 1

   Replace y\to x^r, y'\to rx^{r-1}, y''\to r(r-1)x^{r-2}, then divide out the common factor of x^r to get ar(r-1)+br+c=0.

   Example

   For x^2y''-4xy'+6y=0: x^2\cdot r(r-1)x^{r-2}-4x\cdot rx^{r-1}+6x^r=\big[r(r-1)-4r+6\big]x^r=0, so r(r-1)-4r+6=0, i.e. r^2-5r+6=0.

   Why: every term of a Cauchy-Euler equation collapses to a multiple of the same power x^r — that's what "equidimensional" means — so dividing that common power out leaves a pure algebraic equation in r.

2. 2

   Solve ar(r-1)+br+c=0 for r, factoring if possible.

   Example

   r^2-5r+6=(r-2)(r-3)=0, so r=2 or r=3.

   • Not yet: a repeated or complex pair of roots — covered in Cauchy-Euler with repeated and complex roots

   Why: the roots are exactly the powers of x for which x^r solves the original equation — this section only covers the case where they come out real and distinct.

3. 3

   With distinct real roots r_1\ne r_2, the general solution is y=C_1x^{r_1}+C_2x^{r_2}.

   Example

   y=C_1x^2+C_2x^3

   Why: distinct powers of x are automatically linearly independent on x>0, the same way distinct exponentials were for the constant-coefficient case, so Chapter 6's general-solution theorem applies with nothing further to check.

4. 4

   If you're given y(x_0) and y'(x_0), substitute x_0 into both y and its derivative.

   Example

   For y(1)=5, y'(1)=9: y(1)=C_1+C_2=5, and since y'=2C_1x+3C_2x^2, y'(1)=2C_1+3C_2=9.

   Why: this turns the two free constants into a system pinned down by the actual conditions of the problem, not just the equation.

5. 5

   Solve the resulting 2\times2 linear system for C_1,C_2, then substitute back into y.

   Example

   Substituting C_1=5-C_2 into 2C_1+3C_2=9 gives 10-2C_2+3C_2=9, so C_2=-1 and C_1=6. \boxed{y=6x^2-x^3}

   Why: the general solution is a whole family of curves; the conditions pick out the one curve that actually matches the problem.