Separation of variables hands you an ODE of exactly this shape, almost every time: constant coefficients, second order, set equal to zero. The routine is always the same. Write the characteristic equation. Find its roots. Read the solution's shape straight off them. What changes from one PDE problem to the next is only the sign of one constant in that equation. That sign alone decides whether the roots are real or imaginary.
Guessing y=e^{rx} turns every derivative into the same exponential times a power of r. So the whole differential equation collapses to one polynomial equation in r, with the exponential factored out. Solve that polynomial like any other quadratic. Each root hands back one building-block solution. Which of the three outcomes you get is set by one leftover fact from the quadratic formula. That fact is whether the stuff under the square root is positive, zero, or negative. Real roots grow or decay. A pair of complex roots swings back and forth instead. That's the same rotation-into-cosine-and-sine behavior the last section's Euler's formula described, since that's what e^{i\beta x} does. The zero case sits right at the edge, and it behaves differently again. It needs one extra copy of x tacked on, so the second building block doesn't just copy the first. None of this pins down the two leftover constants sitting in front of each piece. Those get fixed afterward, by whatever starting values or edge values the actual problem hands you.
The characteristic equation of a constant-coefficient linear ODE is an algebraic equation, with no derivatives left in it. You get it by substituting the trial solution y=e^{rx} and cancelling the common exponential factor. Solving it for r is what produces the ODE's building-block solutions. No integration is required. For ay''+by'+cy=0, substitute the trial solution y=e^{rx}. Because y'=re^{rx} and y''=r^2e^{rx}, every term carries the same factor of e^{rx}. That factor is never zero, so it cancels out completely. This collapses the differential equation to exactly one algebraic statement in r: ar^2+br+c=0. Solving this quadratic for its two roots is all that's needed to write down the general solution. No further integration is involved.
The sign of the discriminant b^2-4ac splits the outcome into exactly three cases. Each case hands back a different building-block shape. A positive discriminant gives two distinct real roots r_1,r_2. The solution is y=C_1e^{r_1x}+C_2e^{r_2x}, a sum of two exponentials growing or decaying at different rates. A zero discriminant gives one repeated real root r. The solution is y=(C_1+C_2x)e^{rx}. The extra factor of x is required to keep the second solution independent of the first. Otherwise C_2e^{rx} alone would just be a multiple of C_1e^{rx}. A negative discriminant gives a complex conjugate pair \alpha\pm i\beta. The oscillating solution is y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x). In every case, the two arbitrary constants C_1,C_2 stay unfixed. They get pinned down only once an initial condition or a boundary condition supplies the two further equations needed.
Two real roots, dragged directly
Drag either root along the real axis. Watch y(t) rebuild itself from e^{r_1t} and e^{r_2t} live, and notice what happens to the curve the instant the two roots meet.
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Worked example
Two distinct real roots
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1
Solve y''-y'-6y=0.
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2
The characteristic equation r^2-r-6=0 factors as (r-3)(r+2)=0, giving r=3 and r=-2.
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3\boxed{y = C_1e^{3x} + C_2e^{-2x}}
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Worked example
A repeated root
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1
Solve y''+6y'+9y=0.
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2
The characteristic equation r^2+6r+9=(r+3)^2=0 has the single root r=-3, repeated.
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3\boxed{y = (C_1+C_2x)e^{-3x}}
The extra factor of x is what keeps the second solution independent of the first. Without it, C_2e^{-3x} would just be a multiple of C_1e^{-3x}.
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Where this goes next
The separated spatial equation
Separating a PDE's spatial variable out of the heat, wave, or Laplace equation always produces a constant-coefficient second-order equation, X''+\lambda X=0. Its characteristic equation is r^2+\lambda=0. The sign of \lambda decides everything from here forward. When \lambda>0, you get imaginary roots and oscillating, trigonometric solutions. When \lambda<0, you get real roots and exponential ones. When \lambda=0, you get the repeated-root case. Which sign is even possible depends on the constraint at the domain's edge. That's the entire content of the eigenvalue problem separation of variables sets up. See ch.05, "Separation of variables and the heat equation," where this exact case split first happens for real.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Ordinary Differential Equations
Verify that r=2 is a root of the characteristic equation for y''-y'-2y=0, by substituting. Enter the resulting value (should be 0).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
2^2-2-2=4-2-2=0.
Correct answer
Also accepted:
Why
Substitute r=2 into r^2-r-2.
From Ordinary Differential Equations
Why does substituting y=e^(rx) turn a differential equation into an algebraic one?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
e^(rx) is never zero for any real r -- that's exactly why dividing it out is valid.
The same substitution works at any order, including second.
It's a genuine simplification -- an infinite-dimensional calculus problem collapses to one algebra equation.
Why
Each derivative of e^{rx} pulls out a factor of r, so ay''+by'+cy collapses to (ar^2+br+c)e^{rx} -- an algebraic condition on r, since e^{rx} is never zero.
Compute y', y'' for y=e^(rx) and see what factors out.
From Ordinary Differential Equations
Find the characteristic equation of y''-5y'+6y=0. Enter the expression in r that equals zero.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
r^2-5r+6=0.
Correct answer
Also accepted:
Why
Substitute y=e^(rx) and divide by e^(rx).