Study guide - Math 220 November 28, 2012 1 Exam I 1.1 Linear Equations An equation is linear, if in the form y + p(t)y = q(t). Introducing the integrating factor µ(t) = e p(t)dt the solutions is then in the form y(t) = 1 µ(t) [ ] q(t)µ(t)dt 1.2 Separable Equations Separable equations are in the form y (x) = F (x)g(y) Their solution is found by writing y = F (x) and integrating in x G(y) dy G(y) = F (x)dx. 1
1.3 Exact Equations Consider equations in the form M(x, y)dx + N(x, y)dy = 0, (1) or equivalently y (x) = M(x,y). We say that (1) is exact, if N(x,y) M y = N x. In that case, we are looking for a function ψ, so that { ψx = M(x, y) ψ y = N(x, y) The solutions to (1) are then given by ψ(x, y) = C. 1.4 Homogeneous equations with constant coefficients These are in the form ay + by + cy = 0 We set up the characteristic equation as follows Based on the type of solutions we get for (2), we distinguish several cases ar 2 + br + c = 0. (2) r 1,2 = b ± b 2 4ac 2a 1.4.1 Case I: Two real different roots That is r 1 r 2, both real (that is, if b 2 4ac > 0), we can write the solution to (1) as follows y(t) = C 1 e r1t + C 2 e r2t. 2
1.4.2 Case II: Two complex conjugate solutions If r 1,2 = α ± iβ, (that is, if b 2 4ac < 0), we write y(t) = C 1 e αt cos(βt) + C 2 e αt sin(βt). 1.4.3 Case III: A double root If r 1,2 = r, (if b 2 4ac = 0) y(t) = C 1 e rt + C 2 te rt. 1.5 Inhomogeneous Equations ay + by + cy = g(t). Rules for selecting a particular solution Y 0 : If g(t) = e αt, then seek Y 0 (t) = Ae αt. This is unless α is a root of (2), in which case Y 0 (t) = Ate αt. In the case, where α is a double root of (2), Y 0 (t) = At 2 e αt. If g(t) = e αt cos(βt) or g(t) = e αt sin(βt), then seek Y 0 (t) = Ae αt cos(βt) + Be αt sin(βt), unless α + iβ is a (complex) root of (2). In this case Y 0 (t) = t(ae αt cos(βt) + Be αt sin(βt)). If the right-hand sides discussed above come multiplied by polynomials (say of degree n), modify the corresponding Y 0 by multiplying by a polynomial of the same degree n. 3
2 Exam II 2.1 Variation of parameters Consider the equation y + p(t)y + q(t)y = g(t). (3) Suppose that y 1, y 2 solves the homogeneous equation, y + p(t)y + q(t)y = 0 Then, y(t) = u 1 (t)y 1 (t) + u 2 (t)y 2 (t) solves the original equation, if u 1(t)y 1 (t) + u 2(t)y 2 (t) = 0 u 1(t)y 1(t) + u 2(t)y 2(t) = g(t) (4) That is, solve (4) and then y(t) = u 1 (t)y 1 (t) + u 2 (t)y 2 (t) will be the required solution of (3). 2.2 Mechanical vibrations For a spring-mass system, with a weight w lbs, determine the massfrom m = w g = w 32. If this mass extends the spring L ft., then its coefficient k = w. Then, if γ is its viscosity L coefficient, we write the equation of motion If γ = 0, the solutions of (5) can then be written as mu (t) + γu (t) + ku(t) = 0. (5) u(t) = A cos(ωt) + B sin(ωt) = R cos(ωt δ), where R = A 2 + B 2 is called amplitude, ω = is natural frequency and δ = tan 1 (B/A) is the phase. If γ 0, the solutions are in the form u(t) = Ae γt/2 cos(ωt) + Be γt/2 sin(ωt) = Re γt/2 cos(ωt δ) 4 k m
with 2.3 Laplace transform R = A 2 + B 2, δ = tan 1 (B/A). 4km γ 2 ω =, 2m For a function f and s sufficiently large, we define Some examples are L[f](s) = 0 e st f(t)dt. L[1](s) = 1 s ; L[t](s) = 1 s 2 ; L[eat ](s) = 1 s a, see the list with formulas on page 317 for more info. The usefulness of this method is in the following L[f ](s) = sl[f](s) f(0) This allows us to solve ODE s in the form L[f ](s) = s 2 L[f](s) sf(0) f (0) ay (t) + by (t) + cy(t) = g(t), Indeed, an application of the Laplace transform yields the following equation for F (s) = L[f](s), a(s 2 F (s) sf(0) f (0)) + b(sf (s) f(0)) + cf (s) = L[g](s). Solving for F (s) yields the relation F (s) = L[g](s) + af (0) + (as + b)f(0). as 2 + bs + c Now, if we know how to invert this last explicit formula, we are done. This is done mostly with the methods of the table on page 317, namely formulas 1,2,3, 5,6, 9, 10, 12, 13, 14 come in handy. 5
3 Linear systems of ODE s For linear systems of ODE s in the form x = Ax, a 11 a 12... a 1n a 21 a 22... a 2n where A = is a matrix, we solve as follows. First, determine.... a n1 a n2... a nn the eigenvalues from the characteristic equation a 11 λ a 12... a 1n a 21 a 22 λ... a 2n det.... = 0. a n1 a n2... a nn λ If the eigenvalues λ 1,... λ n are real, find the corresponding eigenvectors by solving the homogeneous system a 11 λ a 12... a 1n a 21 a 22 λ... a 2n.... a n1 a n2... a nn λ The solution is in th form x 1 x 2. x n x(t) = C 1 e λ 1t x (1) +... + C n e λnt x (n) = 0. For each pair of complex conjugate eigenvalues, take λ = α + iβ, find the corresponding eigenvector ξ = a + i b. This will generate the following entries C 1 (e αt cos(βt) a e αt sin(βt) b) + C 2 (e αt cos(βt) b + e αt sin(βt) a) For each repeated eigenvalue λ with a single eigenvector ξ, find the adjoint eigenvector η : (A λi) η = ξ and then write the corresponding solution as x(t) = C 1 e λt ξ + C2 (te λt ξ + e λt η). 6