Differential equations

Second-order linear ODEs with constant coefficients

The general solution of $y'' + ay' + by = f(x)$ is the homogeneous solution plus a particular solution: use the characteristic equation for the first and undetermined coefficients for the second.

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Characteristic equation and general solution

Substituting $y = e^{\lambda x}$ into $y'' + ay' + by = 0$ gives the characteristic equation $\lambda^2 + a\lambda + b = 0$ .

General solution

Distinct real roots $\lambda_1$ , $\lambda_2$ : $y = C_1e^{\lambda_1x} + C_2e^{\lambda_2x}$
Double root $\lambda$ : $y = (C_1 + C_2x)e^{\lambda x}$
Complex roots $\alpha \pm \beta i$ : $y = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$

The initial conditions $y(0)$ and $y'(0)$ give two equations for $C_1$ and $C_2$ .

Undetermined coefficients

Guess the form of a particular solution from the right side $f(x)$ , substitute, and solve for the coefficients.

Trial forms

$f(x)$ a polynomial: a polynomial of the same degree
$f(x) = ke^{cx}$ : $Ae^{cx}$
$f(x) = k\cos\omega x$ or $k\sin\omega x$ : $P\cos\omega x + Q\sin\omega x$
a sum: the sum of the particular solutions (superposition)

Resonance

If $c$ or $\pm\omega i$ is a characteristic root, multiply the trial form by $x$ (by $x^2$ for a double root). For example, $y'' + 4y = \cos 2x$ has a particular solution of the form $x(P\cos 2x + Q\sin 2x)$ .

Euler equations

For $x^2y'' + axy' + by = 0$ , substituting $y = x^{\lambda}$ gives $\lambda(\lambda - 1) + a\lambda + b = 0$ ; with distinct real roots, $y = C_1x^{\lambda_1} + C_2x^{\lambda_2}$ . (The substitution $x = e^t$ turns it into a constant-coefficient equation.)

Q1BasicUniv Y2+Second-order linear ODEs with constant coefficients

Find the solution of the differential equation satisfying $y(0) = 0,\ y'(0) = 6$ .

y'' + 4y' + 13y = 0

Answer

y = 2e^{-2x}\sin 3x

Solution

Solving the characteristic equation,
$\lambda^2 + 4\lambda + 13 = 0 \quad\Rightarrow\quad \lambda = -2 \pm 3i$
The general solution is
$y = e^{-2x}(C_1\cos 3x + C_2\sin 3x)$
From the initial conditions,
$C_1 = 0,\quad -2C_1 + 3C_2 = 6 \quad\Rightarrow\quad C_2 = 2$
Hence
$y = 2e^{-2x}\sin 3x$

Q2StandardUniv Y2+Second-order linear ODEs with constant coefficients

Find the particular solution of the differential equation that is a polynomial.

y'' + 2y' + 2y = -2x^{2} - 2x - 6

Answer

y_p = -x^{2} + x - 3

Solution

Substituting $y = Ax^2 + Bx + C$ and comparing coefficients,
$A = -1,\ B = 1,\ C = -3$
Hence
$y_p = -x^{2} + x - 3$

Q3AdvancedUniv Y2+Second-order linear ODEs with constant coefficients

Find a particular solution of the form $y = Axe^{3x}$ .

y'' - 2y' - 3y = 8e^{3x}

Answer

y_p = 2xe^{3x}

Solution

The characteristic roots are
$\lambda = -1,\ 3$
Substituting $y = Axe^{3x}$ , the $xe^{3x}$ terms cancel and
$4Ae^{3x} = 8e^{3x}$
Hence
$A = 2,\quad y_p = 2xe^{3x}$

Q1BasicUniv Y2+Second-order linear ODEs with constant coefficients

Find the solution of the differential equation satisfying $y(0) = -1,\ y'(0) = 4$ .

y'' + 2y' + y = 0

Q2StandardUniv Y2+Second-order linear ODEs with constant coefficients

Find a particular solution of the form $y = Ae^{2x} + Bx + C$ .

y'' + 4y = 8e^{2x} + 12x + 12

Q3AdvancedUniv Y2+Second-order linear ODEs with constant coefficients

Find the solution of the (Euler-type) equation satisfying $y(1) = 1,\ y'(1) = -5$ .

x^2y'' + xy' - y = 0\quad (x > 0)

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Related topics

First-order differential equations

Eigenvalues, eigenvectors and diagonalization

Differentiation techniques