Math 215/255: Elementary Differential Equations I Harish N Dixit, Department of Mathematics, UBC

Math 215/255: Elementary Differential Equations I Harish N Dixit, Department of Mathematics, UBC

First Order Equations Linear Equations y + p(x)y = q(x) Write the equation in the standard form, Calculate integrating factor, φ = e pdx, Multiply equation with φ, Combine first two terms using product rule Integrate both sides Calculate constant from initial condition Separable Equations Verify if x and y terms can be separated. If yes, Separate x and y terms, f (x)dx = g(y)dy, Integrate both sides Exact Equations M(x, y)dx + N(x, y)dy = 0. Write the equation in the standard form, Verify whether M y = N x Define a function which satisfies ψ ψ x = M and y = N. Integrate the first relation and obtain ψ along with an integration constant h(y) Substitute ψ into the second relation to get h(y) Put everything together. Now ψ=constant is the solution.

Second Order Equations - Homogeneous equations ay + by + cy = 0 Put y = e rt ar 2 + br + c = 0 Real & distinct roots If two roots are r 1 and r 2, then Fundamental solutions: y 1 = e r1t, y 2 = e r2t. General solution: y(t) = c 1 y 1 + c 2 y 2. Complex roots If r = λ ± iµ, then Fundamental solutions: y 1 = e λt cos(µt), y 2 = e λt sin(µt). General solution: y(t) = c 1 y 1 + c 2 y 2. Real & equal roots If r 1 = r 2 = r, then we have only one solution. First solution: y 1 = e rt. Using reduction of order, we try a second solution in the form y 2 = v(t)y 1. Obtain an equation for the variable v (t). Solve for v(t), and this gives y 2 (t). General solution: y(t) = c 1 y 1 + c 2 y 2.

Method of Undetermined coefficients - finding particular solution ay + by + cy = g(t) g(t) differs from y H (t) g(t) same as y H (t) Case I: If g(t) = e αt, then try y p = Ae αt, substitute y p and compare e αt terms on both sides to find A. Case II: If g(t) = cos αt, sin(αt) or a combination of the two, then try y p = A cos αt + B sin(αt), substitute y p and compare sin and cos terms on both sides to find A and B. Case III: If g(t) = t n - a polynomial, then try y p = At n + Bt n 1 +..., substitute y p and compare similar terms on both sides to find the constants. Case I: If g(t) = e αt, then try y p = Ate αt, substitute y p and compare e αt and te αt terms on both sides to find A. Case II: If g(t) = cos αt, sin(αt) or a combination of the two, then try y p = t(a cos αt + B sin(αt)), substitute y p and compare sin, cos, t sin and t cos terms on both sides to find A and B. Case III: If g(t) = t n - a polynomial, then try y p = t(at n + Bt n 1 +...), substitute y p and compare similar terms on both sides to find the constants.

Variation of parameters - finding particular solution y + p(t)y + q(t)y = g(t) If y 1 and y 2 are homogeneous solutions, then y H (t) = c 1 y 1 (t) + c 2 y 2 (t) Particular solution: Put y = u 1 (t)y 1 + u 2 (t)y 2. Obtain y, set u 1 y 1 + u 2 y 2 = 0 - Eq (1). From the simplified y, calculate y to obtain a second equation between u 1 and u 2 - Eq (2). Solve for the variables u 1 and u 2. Integrate these to get u 1 and u 2. Put everything together to get the particular solution.

Laplace Transforms ay + by + cy = g(t) Useful things to remember: F (s) = L{f (t)} = 0 e st f (t)dt. If L{y(t)} = Y (s), then L{y (t)} = sy (s) y(t = 0) and L{y (t)} = s 2 Y (s) sy(t = 0) y (t = 0). Never mix up variables s and t in the same equation. Unit step function: { 0 0 t < c, u c (t) = 1 t c. If f (t) = L 1 {F (s)}, then L 1 {e cs F (s)} = u c (t)f (t c). This is available in the table. Where ever required, express g(t) in terms of u c (t) and then modify the expression to get it in the form u c (t)g(t c). Always check to make sure your expression for g(t) is correct.

Homogeneous Linear Equations X = AX Put X = ξe λt (A λi)ξ = 0 λ s: real & distinct Real & distinct eigenvalues Write two fundamental solutions as X (1) = ξ (1) e λ1t and X (2) = ξ (2) e λ2t General solution is X = c 1 X (1) + c 2 X (2). Eigenvalues: A λi = 0 Complex eigenvalues λ s: complex Write X (1) = ξ (1) e λ+t Take real and imaginary part of X (1) If U = Re{X (1) } and V = Im{X (1) }, then general solution is X = c 1 U + c 2 V. λ s: repeated Repeated eigenvalues First solution: X (1) = ξ (1) e λt Second solution: Put X = ξte λt + ηe λt Substitute to obtain (A λi)η = ξ and find η. Put everything together to get X (2). General solution is X = c 1 X (1) + c 2 X (2).

Homogeneous Linear Equations: Classification of critical point X = AX (0, 0) is a critical/equilibrium point Steps: Put X = ξe λt. Substitute to get (A λi)ξ = 0. Compute eigenvalues and eigenvectors of the matrix A. If λ 1 and λ 2 are real and of same sign - Node. If λ1 and λ 2 are real and of opposite sign - Saddle. If λ1 and λ 2 are complex conjugates - Spiral. The real part of λ 1,2 controls the growth/decay of spiral. If λ1 and λ 2 are purely imaginary, i.e. no real part - Center. If Re{λ 1,2 } > 0, then the critical point is unstable. Two solutions: X (1) = ξ (1) e λ1t and X (2) = ξ (2) e λ2t

Nonlinear system of equations x = F (x, y), y = G(x, y). Find critical points F (x, y) = 0, G(x, y) = 0 at (x 0, y 0 ) J = Construct Jacobian ( ) Fx F y G x G y Real eigenvalues λ s: real Nodes and Saddles Near each (x 0, y 0 ) Solve U = JU λ s: imaginary Purely imaginary eigenvalues Centers λ s: complex conjugates Complex eigenvalues Spirals Global phase portrait: Putting everything together Draw all the local phase portraits separately Use common sense to connect all the arrows