Guides / Differential equations

Second-order linear differential equations: the characteristic roots decide the form, and the constants come last

An equation ay″ + by′ + cy = f(t) with constant a, b and c describes anything with inertia and a restoring force: a mass on a spring, an electrical circuit, a swinging pendulum. Trying y = e^rt turns the differential equation into a quadratic for r, and the kind of roots it has decides what the solution looks like. A forcing term f(t) adds one particular solution on top. The mistakes come from a repeated root given two copies of the same exponential, a guess for the particular solution that already solves the homogeneous equation, and initial conditions applied before the particular solution is in.

Why it works

Every derivative of e^rt is a multiple of itself: (e^rt)′ = re^rt and (e^rt)″ = r²e^rt. Put into ay″ + by′ + cy = 0, the left side becomes (ar² + br + c)e^rt, and since e^rt is never 0, it vanishes exactly when r is a root. The equation is linear, so any combination of solutions is a solution, and a second-order equation needs two independent ones to meet two initial conditions. A repeated root supplies only one exponential, and te^rt turns out to be the second. Complex roots give e^{(α ± βi)t} = e^αt(cos βt ± i sin βt), and combining the two leaves the real solutions with cosine and sine.

For a forcing term, if y_p is any one solution, every other solution differs from it by a solution of the homogeneous equation. So y = y_h + y_p is the general solution, and its two constants are still free until the initial conditions fix them.

Example 1: two distinct real roots

Solve y″ − 5y′ + 6y = 0 with y(0) = 2 and y′(0) = 5.

1. r² − 5r + 6 = (r − 2)(r − 3) = 0The roots are 2 and 3.
2. y = c₁e^2t + c₂e^3tOne exponential for each root.
3. c₁ + c₂ = 2, 2c₁ + 3c₂ = 5y(0) and y′(0), with y′ = 2c₁e^2t + 3c₂e^3t.
4. y = e^2t + e^3tSubtracting twice the first equation from the second gives c₂ = 1, so c₁ = 1.

Example 2: a repeated root

Solve y″ − 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 4.

1. r² − 4r + 4 = (r − 2)²One root, r = 2, twice. The discriminant b² − 4ac is 16 − 16 = 0.
2. y = (c₁ + c₂t)e^2tThe second solution is te^2t.
3. y′ = (2c₁ + c₂ + 2c₂t)e^2tProduct rule on c₂te^2t.
4. c₁ = 1, 2 + c₂ = 4 ⇒ y = (1 + 2t)e^2ty(0) = c₁ and y′(0) = 2c₁ + c₂.

Example 3: complex roots

Solve y″ + 4y′ + 13y = 0 with y(0) = 2 and y′(0) = 0.

1. r = (−4 ± √(16 − 52))/2 = −2 ± 3iThe discriminant is −36, so the roots are complex: α = −2 and β = 3.
2. y = e^−2t(c₁ cos 3t + c₂ sin 3t)An oscillation with angular frequency 3, inside an envelope that decays like e^−2t.
3. y(0) = c₁ = 2, y′(0) = −2c₁ + 3c₂ = 0At t = 0 the derivative of e^−2t contributes −2c₁ and the derivative of sin 3t contributes 3c₂.
4. y = e^−2t(2 cos 3t + 43 sin 3t)c₂ = 4/3. A positive coefficient on y′ means α < 0: damping makes it die out.

Example 4: a forcing term, with the constants found last

Solve y″ + 4y′ + 3y = 9t with y(0) = 0 and y′(0) = 0.

1. r² + 4r + 3 = (r + 1)(r + 3) ⇒ y_h = c₁e^−t + c₂e^−3tThe homogeneous solution, with its constants left free.
2. y_p = At + B: 4A + 3(At + B) = 9tThe input is a degree-1 polynomial, so guess one. y_p″ = 0 and y_p′ = A.
3. 3A = 9, 4A + 3B = 0 ⇒ y_p = 3t − 4Match the t terms, then the constants.
4. c₁ + c₂ − 4 = 0, −c₁ − 3c₂ + 3 = 0Now apply y(0) = 0 and y′(0) = 0 to y_h + y_p, the whole solution.
5. y = 92e^−t − 12e^−3t + 3t − 4Adding the equations gives −2c₂ − 1 = 0, so c₂ = −½ and c₁ = 9/2. Check: y(0) = 9/2 − 1/2 − 4 = 0.

Common mistakes, and the exact line they happen on

1. Two copies of one exponential for a repeated root

The equation from Example 2.

1. (r − 2)² = 0, r = 2 twiceRight so far.
2. y = c₁e^2t + c₂e^2tThis is the mistake. This is (c₁ + c₂)e^2t, one constant times one function. y(0) = 1 makes it e^2t, whose y′(0) is 2, and no choice of constants reaches the required 4. The second solution is te^2t.

2. A guess that already solves the homogeneous equation

Find a particular solution of y″ − 3y′ + 2y = e^2t.

1. r² − 3r + 2 = (r − 1)(r − 2)Right so far: the roots are 1 and 2.
2. y_p = Ae^2t: A(4 − 6 + 2)e^2t = e^2t, so 0 = 1This is the mistake. e^2t is part of y_h, so it makes the left side 0 whatever A is. The equation has a particular solution; the guess was in the wrong family. Multiply by t: y_p = Ate^2t gives A = 1, so y_p = te^2t.

3. Initial conditions applied to y_h alone

The equation from Example 4.

1. y_h = c₁e^−t + c₂e^−3t, y_p = 3t − 4Right so far.
2. c₁ + c₂ = 0, −c₁ − 3c₂ = 0 ⇒ y = 3t − 4This is the mistake. The initial conditions belong to y, not to y_h. This answer solves the differential equation, but y(0) = −4 and y′(0) = 3, not 0 and 0. Add y_p first, then solve for the constants: c₁ = 9/2 and c₂ = −½.

Practice

1. Solve y″ − y = 0 with y(0) = 3 and y′(0) = 1.

   Show answer

   r = ±1, so y = c₁e^t + c₂e^−t with c₁ + c₂ = 3 and c₁ − c₂ = 1: y = 2e^t + e^−t.

2. Solve y″ + 2y′ + y = 0 with y(0) = 0 and y′(0) = 3.

   Show answer

   r = −1 twice, so y = (c₁ + c₂t)e^−t. c₁ = 0 and c₂ = 3: y = 3te^−t.

3. Solve y″ + 9y = 0 with y(0) = 1 and y′(0) = 6. What are the amplitude and period?

   Show answer

   r = ±3i, so y = cos 3t + 2 sin 3t. The amplitude is √(1² + 2²) = √5 and the period is 2π/3.

4. Find a particular solution of y″ + y = 3 cos 2t.

   Show answer

   Try A cos 2t + B sin 2t: −4A + A = 3 and −4B + B = 0, so y_p = −cos 2t. The guess is safe because 2i is not a root of r² + 1.

5. Find a particular solution of y″ + 2y′ − 3y = 8e^t.

   Show answer

   The roots are 1 and −3, so e^t solves the homogeneous equation and the guess is Ate^t. Substituting gives 4Ae^t = 8e^t, so y_p = 2te^t.