Recipe

Solve a Cauchy-Euler equation

Use when: you have ax^2y''+bxy'+cy=0 (x>0), and the characteristic equation's roots might be repeated or complex, not just distinct and real

Goal: the general, real-valued solution y(x), covering every possible root outcome

Taught in Cauchy-Euler with repeated and complex roots · extends Solve a Cauchy-Euler equation

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

   Why: every term of a Cauchy-Euler equation collapses to a multiple of the same power x^r, so dividing that common power out leaves a pure algebraic equation in r — this holds regardless of what kind of roots that equation turns out to have.

2. 2

   Solve ar(r-1)+br+c=0 for r, and check its discriminant before you finish factoring.

   Example

   r^2+4r+4=(r+2)^2=0, a repeated root r=-2.

   • distinct real roots r_1\ne r_2 — use y=C_1x^{r_1}+C_2x^{r_2} below.
   • a repeated real root r (discriminant zero) — use y=(C_1+C_2\ln x)x^{r} below.
   • a complex conjugate pair r=\alpha\pm\beta i — use y=x^{\alpha}\big(C_1\cos(\beta\ln x)+C_2\sin(\beta\ln x)\big) below.

   Why: the roots are exactly the powers of x for which x^r solves the original equation, and the discriminant of the resulting quadratic in r decides which of the three cases you're in — there is no fourth case.

3. 3

   Distinct roots give y=C_1x^{r_1}+C_2x^{r_2}; a repeated root r gives y=(C_1+C_2\ln x)x^{r}; a complex pair \alpha\pm\beta i gives y=x^{\alpha}\big(C_1\cos(\beta\ln x)+C_2\sin(\beta\ln x)\big).

   Example

   With the repeated root r=-2: y=(C_1+C_2\ln x)x^{-2}.

   Why: a repeated root only hands over one genuine solution, x^r — reduction of order produces the missing second solution x^r\ln x, and a complex pair's raw complex powers split into real cosine and sine pieces the same way Euler's formula did for the constant-coefficient case, via the substitution t=\ln x this section's technical block derives.

4. 4

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

   Example

   For y(1)=3, y'(1)=1: y(1)=C_1=3. Since y'=\big[C_2-2(C_1+C_2\ln x)\big]x^{-3}, y'(1)=C_2-2C_1=1.

   Why: this turns the 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

   With C_1=3: C_2-6=1, so C_2=7. \boxed{y=(3+7\ln x)x^{-2}}

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