Week 8 Second-order ODEs Second-order linear homogeneous ODEs Second-order linear non-homogeneous ODEs Method of undetermined coefficients Method of variation of parameters Reduction of order Eigenvalue problems for ODEs Modelling: Forced oscillations

Second-order linear homogeneous ODEs The general form of a 2nd-order linear ODE is (1) ۞ If r(x) = 0, Eq. (1) is called homogeneous. Otherwise, (1) is said to be non-homogeneous.

Theorem 1: The Superposition Principle Let y1(x) and y2(x) be particular solutions of a 2nd-order linear homogeneous ODE, and let them be independent, i.e. Then the general solution of this ODE is where C1 and C2 are arbitrary constants. Comment: Theorem 1 can be used only if we have somehow found y1,2(x).

Consider a particular case: of 2nd-order linear homogeneous ODEs with constant coefficients, i.e. (2) where p and q are constants. An educated guess: linear homogeneous equations (of arbitrary order) with constant coefficients admit exponential solutions, i.e. those of the form (3) where μ is a constant. To find μ, substitute (3) into (2): hence...

(4) ۞ (4) is called the characteristic equation [of Eq. (2)]. Eq. (4) is a quadratic equation for μ. Its solutions are There are 3 possible cases: (1) real distinct roots (p2 > 4q), (2) complex roots (p2 < 4q), (3) real double root (p2 = 4q).

Case 1: Since μ1 ≠ μ2, two independent (why Case 1: Since μ1 ≠ μ2, two independent (why?) particular solutions exist, and, according to The Superposition Principle, the general solution is Case 2: If the roots of the characteristic equation are complex, the general solution is, generally, also complex, (5)

One can make solution (5) real by an appropriate choice of c1,2. Indeed, rearrange (5) in the form hence, hence, using Euler’s formula, hence, where C1 = c1 + i c2 and C2 = c1 – i c2 can be made real (how?).

Case 3: If μ1 = μ2 = μ, we have only one particular solution and, thus, can’t use The Superposition Principle, It turns out that the second particular solution can be simply guessed (and verified by substitution, of course): so the general solution is

2nd-order linear homogeneous ODEs with constant coefficients (summary) Given write the characteristic equation, and solve it,

Case 1: Real distinct roots (μ1 ≠ μ2, Im μ1,2 = 0): Case 2: Complex roots (μ1,2 = α ± iω): Case 3: Real double root (μ1 = μ2 = μ):

Example 1: Solve the following ODEs: The answers:

Using the general solution, we can solve any initial-value problem. Note, however, that the general solution of a second-order ODE involves two arbitrary constants – so, to fix them both, one needs two initial conditions (one for y, plus another one for y' ). Example 2: Solve the following initial-value problem: The answer:

In initial-value problems, the initial conditions are set for the same value of x (typically, x = 0). ۞ A 2nd-order ODE + two additional conditions set for different values of x constitute a boundary-value problem. Example 3: Solve the following boundary-value problem: The answer:

2. Second-order linear non-homogeneous ODEs Some 2nd-order linear ODEs with variable coefficients, (6) can be solved using the following theorem. Theorem 2: The general solution of Eq. (6) can be represented in the form where yp is a particular solution of Eq. (6) and yh is the general solution of the corresponding homogeneous ODE, i.e.

Example 4: Solve Solution: Step 1: Solve the corresponding homogeneous ODE, Step 2: Guess a particular solution of the ‘full’ (non-homogeneous) ODE, Step 3: Use Theorem 2, Q: How do we find yp in more complicated cases?...

3. Method of undetermined coefficients Consider (7) If p and q are constants, and r(x) is one of the following functions: or a linear combination of the same. Then, yp for Eq. (7) can be guessed: generally, you assume it to be from the same class of functions as r(x).

Example 5: Find a particular solution of The answer: We can also guess yp in some other cases...

r(x) yp k sin ωx A cos ωx + B sin ωx k cos ωx k eγx A eγx k xn A0 + A1 x + ... + An xn Observe that, for r = sin... and r = cos..., the particular solution yp includes both the sine and cosine. Also, if r = xn, then yp includes xn and the lower powers of x as well.

Warning 1 (Modification Rule 1): The table on the previous slide works only if r(x) is not a solution of the corresponding homogeneous ODE. If it is, yp from the table should be multiplied by x. Warning 2 (Modification Rule 2): In some cases, the modified yp is still a solution of the corresponding homogeneous ODE. If it is, the modified yp should be multiplied by another x (i.e. yp from the table should be multiplied by x2). Comment: MR2 is typically needed if r = eγx and the corresponding homogeneous ODE belongs to Case 3 (in which case eγx and x eγx are both solutions).

Example 6: Find a particular solution for (8) Solution: Let’s first try the ‘basic rule’ of the method of undetermined coefficients, hence, hence, Eq. (8) becomes which can’t be satisfied by any choice of A and B.

The basic rule of the method of undetermined coefficients failed for Eq. (8) because its r.-h.s. r(x) is a solution of the corresponding homogeneous ODE (verify this by substitution). Next, let’s try MR2: hence, hence (skipping some straightforward algebra),