Second-Order Linear ODEs #

Definition: Linear Second Order Differential Equation: y’’ + p(x)y’ + q(x)y = r(x)

It’s called linear because everything is of the first order (power <= 1) and not of some nonlinear function like sin or cos.

Definition: Homogeneous: 2nd order ODE with r(x)=0

Definition: Nonhomogeneous: 2nd order ODE with r(x) not = 0

f(\lambda x,\lambda y) = \lambda^n f(x,y)

Homogeneous ODEs don’t need to be second order.

Fundamental Theorem for the Homogeneous Linear ODE #

For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions

We have two initial conditions in second-order homogeneous linear ODEs. These are used to set the constants for the linear combination in \(y = c_1y_1 + c_2y_2 \).

Definition: General Solution

Definition: Basis: y1 and y2

Definition: Particular Solution: solution with constants non-arbitrary

7/29

graph TD
    Wall[Fixed Wall] --> Spring(Spring)
    Spring --> Mass(Mass)

Undamped mass-spring system. y’’ + k/m y = 0 k is the spring constant. m is the mass underneath the spring system.

from mg = ks_0 k = mg/s_0 \star

General solution: y = Acos omega t + B sin omega t \star where omega = sqrt(k/m) \star

frequency (ideal) = omega_0 / 2pi

Damped mass-spring system. It’s the same system with added resistance.

F_2 is the damping force. F_2 = -cy' y is displacement. c is the damping constant.

ma = F1 + F2 = spring force + damping force

my’’ = -ky - cy'

some algebra to get y’’ + (c/m)y’ + (k/m)y = 0

using the sections before, we find the characteristic equation: \(\lambda^2 + \frac{c}{m}\lambda + \frac{k}{m} = 0\) Using the quadratic formula, \(\lambda = -\frac{c}{2m} \pm \frac{1}{2m}\sqrt{c^2-4km} = \alpha \pm \beta\)

when c^2 - 4km > 0, it’s an over-damped system.

Cases:

Case 1: distinct real roots, overdamping

y = c_1e^{{-\alpha + \beta}x} + c_2e^{{-\alpha - \beta}x}

Case 2: two = real roots \lambda_1 = \lambda_2 = -\alpha = -c/2m critical damping

y = (c_1+c_2x)e^{{-c/2m}x}

Case 3: Complex roots \lambda = -\alpha \pm \frac{1}{2m}\sqrt{4km - c^2}i lol he just flipped the root this is an underdamping case.

y = e^{{-c/2m}x}(A cos omega x + B sin omega x) omega = 1/2m * \sqrt{2km-c^2}

Discussion of 3 cases. case 1: c^2 > 4km as the limit of the ind variable goes to infinity, \alpha is 1/2m \sqrt{c^2} okay, \beta is 1/2m {c^2-4mk} so, the first exponential goes to 0.

Overdamping comes from the exponential decay and both decaying.

Case 2: the limit goes to 0, but we have t in the expression, so we use l-hopital and we can solve it nicely. \(\lim_{t \to \inf} (y) = \lim_{t \to \inf} (c_1 e^{\frac{-c}{2m}t} + c_2 t e^{\frac{-c}{2m}t})\)

Case 3: Briefly, Acos(anything) + Bsin(anything) \le A(1)+B(1) Now, limit of e^{-asldfjl} * (A+B) is (A+B)*limit(lajsdf) = (A+B)*0

Quick Summary,

Example 2. Similar to example 1. but with a damping force.

Givens: m = 10 kg k = 90

and try three values of c, namely, c_1 = 100 kg/sec c_2 = 60 kg/sec c_3 = 10 kg/sec

You can calculate the determinate? what’s it called for the quadratic formula.

y’’ + 10y’ + 9y = 0 \lambda^2 + 10\lambda + 9 = 0 now, lambda_1 = -1 , \lambda_2 = -9 y = c1e^-t + c_2e^-9t

2nd y’’ + 6y’ + 9y = 0 do the algebra in latex. y = (c1+c_2t)e^-3t

3rd y’’ + y’ + 9y = 0 do the lambda shit. the discriminant is negative. lambda = {-1 \pm \sqrt{35}i}/2 = -1/2 \pm sqrt{35}i/2

y = e^{-0.5t}(A cos \sqrt{35}t/2 + B sin \sqrt{35}t/2)

2.5

Euler-Cauchy Equations x^2y’’ + axy’ + by = 0

substitute y = x^m and we solve for m. y’’ = m(m-1)x^(m-2) y’ = mx^(m-1)

plug in and do the algebra. m^2 + (a-1)m + b = 0 now, we have a quadratic.

Now, we have cases:

Real distinct roots. y_1 = x^m1, y_2 = x^m2

check y1/y2 (unclear) = x^{m1-m2} is nonconstant bc m1-m2 != 0.

y = c1y1 + c2y2 = c1x^m1 + x2x^m2

Case 2: Double(duplicate) real root. m1 = m2 = (1-a)/2

y1 = x^m = x^{(1-a)/2}

for y2, we use reduction of order. Let y = y2 = uy1 y’ = u’y1 + uy1' y’’ = u’‘y1 + 2u’y’ + uy’'

We plug in and let U = u’. u = integral Udx where U = 1/y1^2 e^(- \int P(x)dx) now, U = 1/y1^2 1/x^a U = 1/((x^((1-a)/2))^2) 1/x^a = 1/x u = integral of Udx = ln|x| we drop the abs value.

y2 = uy1 = lnx y1

so, y = (c1+c2lnx)x^{(1-a)/2} \star

Ex 1: Solve x^2 y’’ - 5xy’ + 9y = 0 you can do this. double real .

Case 3: not covered by test 2 complex roots.