Recipe

Solve a homogeneous second-order linear ODE

Use when: you have ay''+by'+cy=0 with constant coefficients, and the characteristic equation's roots are real — possibly repeated

Goal: the general solution y(x), whether the roots are distinct or repeated

Taught in Repeated real roots · extends Solve a homogeneous second-order linear ODE · a later section extends this recipe with more cases

1. 1

   Replace y''\to r^2, y'\to r, and y\to1 to get ar^2+br+c=0.

   Example

   For y''-6y'+9y=0 (a=1,b=-6,c=9): r^2-6r+9=0.

   Why: substituting the trial solution y=e^{rx} into the ODE and dividing out e^{rx} — never zero — collapses the differential equation to this single algebraic one.

2. 2

   Solve ar^2+br+c=0 for r; the discriminant b^2-4ac tells you which case you're in before you even factor.

   Example

   b^2-4ac=(-6)^2-4(1)(9)=36-36=0 — the repeated-root case. Factoring confirms it: r^2-6r+9=(r-3)^2=0, so r=3.

   • b^2-4ac>0, two distinct real roots r_1\ne r_2 — use y=C_1e^{r_1x}+C_2e^{r_2x} below.
   • b^2-4ac=0, one repeated real root r — use y=(C_1+C_2x)e^{rx} below.
   • Not yet: b^2-4ac<0, a pair of complex conjugate roots — covered in Complex roots and Euler's formula

   Why: the characteristic equation's roots are exactly the values of r for which e^{rx} solves the ODE, and the discriminant's sign is what decides which of the two real-root cases you're in.

3. 3

   Distinct roots give y=C_1e^{r_1x}+C_2e^{r_2x}; a repeated root r gives y=(C_1+C_2x)e^{rx}.

   Example

   With the repeated root r=3: y=(C_1+C_2x)e^{3x}.

   Why: a repeated root only hands over one genuine exponential solution, e^{rx} — reduction of order (Chapter 6, section 10) is what produces the missing second solution xe^{rx}, never a second copy of e^{rx}.

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(0)=2, y'(0)=1: y(0)=C_1=2. Since y'=C_2e^{3x}+3(C_1+C_2x)e^{3x}, y'(0)=C_2+3C_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=2: C_2+3(2)=1, so C_2=-5. \boxed{y=(2-5x)e^{3x}}

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