Differential Equations

Laplace Transform

- an integral transform that is widely used to solve linear differential equations with constant coefficients 1. Given a table of Laplace Transforms 2. Take the Laplace transform of both sides of the equation to get an algebraic equation - use the properties of the Laplace transform. we can transform a constant coefficient differential equation into an algebraic equation 3. Solve for F(s) - you may need to use partial fraction decomposition 4. Write out the solution in physical space and apply L^-1 to retrieve the solution - use the given Laplace transform table

Variation of Parameters

1. Consider the differential equation y"+q(t)y'+r(t)y=g(t) 2. Assume that y1(t) and y2(t) are a fundamental set of solutions for y"+q(t)y'+r(t)y'=0 3. Then a particular solution the non-homogeneous differential equation is yp = -y1 $y2g(t)/W(y1,y2) dt + $ y1g(t)/W(y1,y2) dt 1. solve for homogeneous equation and get y1 and y2 2. Set up a system of linear equations in the form of a matrix 3. [u1' u2'] = 1/det ( inverse wronskian) [ 0 g(t)] 4. Solve the matrix for u1' and u2' 5. Integrate to obtain u1 and u2 6. yp = u1y1 + u2y2 and yg = yp + yo

W(y1,y2)(t)=0 Wronskian

1. Differentiate and plug in the initial conditions yo = y(to) = C1y1 (to) + C2y2(to) y'o = y'(to) = C1y'1(to) + C2y'2(to) 2. Use Cramer's rule: C1= |yo y2(to)| C2=|y1(to) yo| |y't y'2(to)| |y't(to) y'o| __________________ ------------------- |y1(to) y2(to)| |y1(to) y2(to)| | y'1(to) y'2(t0)| |y'1(t0) y'2(to)| Where |a b| = ad-bc |c d| The denominator is known as the Wronskien The two solutions are called a fundamental set of solutions and the general solution is y(t) = c1y1(t) + C2y2(t)

Second order mechanics

1. Draw the free body diagram 2. Sum the forces acting on the mass (keep track of the reference direction associated with theses forces). Through Newton's second law the sum of these forces is equal to the mass times acceleration -Fb - Fk + -b dx/dt -kx = md2x/dt2 3. Rearrange the equation in standard form m d2x/dt2 + bdx/dt - kx = 0 which is a second order homogeneous equation 4. Initial conditions x(o)= xo - initial displacement v(o) = vo - initial velocity

homogeneous equations

1. Recognize that your equation is an homogeneous equation by checking that f(tx,ty) = f(x,y) meaning that f(tx,ty) is independent of the variable t -> write the equation so that it is terms of y/x 2. Write out the substitution z = y/x or v = y/x 3. Through easy differentiation. find the new equation satisfied by the new function form: x dz/dx + z = f(1,z) y = xv -> v + xv' = f in terms of v in replace for y/x 4. solve the new equation by separation to find z or v 5. Go back to the old function y through substitution y = xz - substitute v = y/x back in and solve for y

Second order Electrical

1. draw the electrical diagram 2. recall the constitutive relationships vR = IR vL = L * I' vC= Q/C 3. Construct the equation E(t) = IR + 1/C Q + LI' or LQ" + RQ' + 1/C Q = E(T) which is a second order homogeneous equation NOTE: If the elements are in a parallel circuit then E(t) = 0

Finding l and u

1. understand determinants 2. write our the eigenvalue equation 3. set up the characteristic equation 4. Obtain the characteristic polynomial 5. Solve the characteristic polynomial for the eigenvalues 6. substitute the eigenvalues into the eigenvalue equation, one by one 7. row-reduce the resulting matrix 8. obtain the basis for the eigenspace

Linear systems

A mathematical model of a system based on the use of a linear operator an eigenvector or characteristic vector is a non-zero vector that changes by only a scaler factor when that linear transformation is applied to it an eigenvalue or characteristic value is the characteristic root associated with the eigenvalue

Equilibrium solution

A solution to a DE whose derivative is zero everywhere. Come in two flavours stable and unstable 1. stable - one that other solutions are trying to get to 2. unstable - one that the other solutions are trying to get away from - saddle, sink node, source node, spiral sink, spiral source, center

exact equations

An equation M(x,y) dx + N(x,y) dy = 0 is "exact" if qM/qy = qN/qx . In this case M = qf/qx N= qf/qy and the equation becomes qf/qx dx + qf/qy dy = 0. df(x,y) = 0 f(x,y) = C or if possible y = g(x) 1. check to see if the equation is exact or not - calculate the partial derivatives (M)dx + (N)dy = 0 qM/qy qN/qx 2. Integrate P with respect of x to obtain F F= &qf/qx Integrate M with respect to x = &(M)dx = Integrate + constant with respect to y called R(y) 3. Take the derivative of F with respect to y to find R'(y) qF/qy = (N)dy Integrate N with respect to y 4. Find the constant of integration &R'(y) = R(y) Integrate the constant of integration 5. Arrive at the solution - Plug in the consant

Convolution

An integral that expresses the amount of overlap of one function g as it is shifted over another function f. These functions are usually piecewise Defined as a product of functions f and g that are objects in the algebra of Schwartz functions in R^n. Convolution of two functions f and g over a finite range [0,t] is given by [f * g](t)= INT T->0 f(T)g(t-T)dT where the symbol [f * g] denotes convolution of f and g

integrating factor

Assume Mdx + Ndy = 0 not exact Multiply by u(integrating factor): (Mu)dx + (Nu)dy = 0 (now can claim that the equation is exact) (Mu)y = (Nu)x My * u + M *uy = Nx * u + N * ux Two cases 1) u = ux (depends on x only) My * u = Nx * u + N * ux ux = (My - Nx)u/ N u'(x) = My - Nx/N * u 2) u = u(y) (depends on y only) My * u + M * uy = Nx *u uy = (Nx - My)u/M u'(y) = Nx - My/ M * u 1. Find out if the equation is exact by taking partial derivative If Not 1. There exists a function u(x,y) in which you can multiply the equation by to make the equation exact 2. There are two cases on how to find u(x,y) Case 1: Mdx + Ndy = 0 Not exact: My with not equal Nx: My - Nx =/ ) however if My-Nx/N is a function of x only, let it be denoted by {(x)} then u(x) = e^${(x)}dx will be the integrating factor Case 2: If My-Nx/-M is a function of y only, let it be denoted by Y(y) then u(y) = e^$Y(y)dy will be the integrating factor

charging and discharging a capacitor

C = capacitor q = charge V = voltage V = 1/C q W(work) = 1/2 CV^2 Charging a capacitor in RC - ohms law: V=RI V= 1/C q I= dq/dt E = Vr + Vc E = RI + 1/C q E = R dq/dt + 1/C q (linear equation) -> dq/dt + 1/RC q = E/R Integrate: q(t)= E * C + k e^-t/RC q(t) = qf (1 - e^-t/RC) limit is EC Discharging an capacitor in RC E= 0 R dq/dt + 1/C q = 0 dq/dt + 1/RC q = 0 q(t) = e^-t/RC * K q(t) = qmax e^-t/RC Charging 1. Let I = dQ/dt E - dQ/dt R - Q/C = 0 2. The above equation is a linear equation that can be solved for the function of time Q(t) = Qo [ 1-e^-t/RC] where Qo = CE 3. Differentiate this function to get the current as a fuction of time I(t) = (Qo/RC)e^-t/RC = Ioe^-t/RC where Io = E/R is the max current possible Discharging 1. Let I = dQ/dt. Rewrite the equation R(dQ/dt) = - Q/C 2. Linear equation that can be solved for Q as the function of time Q(t) = Qoe^-t/RC were Qo is the initial charge on the capicator 3. Differentiate the expression to get the current as a function of time I(t) = - (Qo/RC)e^-t/RC = -Ioe^-t/RC where Io = Qo/RC

2nd order linear homogeneous diff eq

Form: ay"+by'+cy=0 where a, b and c are constants and a=/ 0 1. use the fact that the exponential function y=e^rx (where r is a constant) has the property that its derivative is a constant multiple of itself 2. Substitute these expressions into equation 3. Take out e^rx 4. Find the roots r1 and r2 by either factoring or using the quadratic formula (there are 3 cases) Case 1: b^2 - 4ac > 0 the roots are real and distinct numbers. Case 2: b^2-4ac=0 r=-b/2a. Case 3: b^2-4ac (the roots are complex numbers written in the form a +_ bi we write the solution as y= C1e^(a+ib)x+C2e^(a-ib)x or y=e^ax(C1 cos(bx)+C2sin(bx)) We can actually solve these problems using initial conditions 1. find the general solution using the techniques above (know that the general solution is y(x)=C1e^r1x + C2e^r2x 2. Differentiate the equation to get y'(x) 3. plug in the conditions so you are just left with C terms 4. Solve using linear combination algebra techniques to solve for C1 and C2. (Get one and plug in to find the other)

Euler's Equations

ax^2y" + bxy"+cy=0 Assume x>0 1. Plug y(x)=x^r into the differential equation 2. ar(r-1)+b(r)+c=0 (characteristic equation) 3. Solve for r -> 3 cases real, double, and complex Real: General solution is y(x)=C1x^r+C2x^r Double: get single solution like last time but we need 2 1. y2(x)=x^rlnx 2. plug in and rewrite x^r(C1 +C2 lnx) Complex: Solve the quadratic using the quadratic formula r1,2 = a +_ bi 1. use the first root x^a+bi 2. recall that x^r = e^lnx^r = e^r lnx 3. Plug in the root gives us = x^a (C1 cos(b lnx) + C2 sin(b lnx)

Abel's Theorem

If y1 and y2 are 2 solutions to the equation y" + p(x)y'+q(x)y=0 Then W(y1,y2)= Ce^-$p(t)dt

the case of AC for RC circuits and LC circuits

LR Circuit: Vl + Vr = E L dI/dt + RI = E I'(t) + R/L * I = E/L (linear equation) I(t) = Imax (1-e^-R/L*t) AC Circuit: E = V0 coswt LI' + IR = V0 coswt I' + I/L R = V0/L coswt (linear) I(t) = V0/L e^-R/L t int(e^R/L s cosws ds) = V0/sqrt(R^2+(wl)2) *cos(wt-p) - V0R/R^2(wL)^2 * e^-R/Lt

Wronskian

Let y = y₁(x) and y = y₂(x) be solutions of a second order linear homogenous differential equation. The function W defined by W [y₁,y₂](x) = = y₁(x) y'₂(x) - y₂(x) y'₁(x) determinant of y₁(x) y₂(x) y'₁(x) y'₂(x) Note: y₁ and y₂ are linearly independent i.f.f. W[y₁,y₂] ≠ 0

Reduction of order

Method of solving 2nd order differentials of the form p(t)y" + q(t)y' + r(t) y = g(t) Note: first solution is given y1(t) = ? 1. Second solution is of the form y2(t) = y1(t)v . Find the first and second derivatives 2. Plug the derivatives into the equation for y", y', and y 3. Rearrange and simplify 4. Use variable transformation w = v' -> w' = v" and substitute 5. Equation is now an first order differential equation. Solve for v 6. Plug this v for y2 and then you have the full solution y = c1y1 + c2y2

Case of real repeated eigenvalues

Need to come up with a second solution. Recall the double root case with second order ODE. In that case we simply add a t to the solution and we are able to get a second solution. x2= te^lt N + e^lt P is a solution provided P is a solution to (A-LI)P= N where P is an unknown vector that we need to determine 1. find the eigenvalues for the system 2. find the eigenvector for this eigenvalue 3. find P by solving for it AP=N 4. We can now write down the general solution to the system

Convolutions and Laplace Transform

Use the property of convolution integrals that (f * g) (t) = ( g * f)(t) Main fact allows us to take the inverse transform of a product of transforms L{f * g}= F(s)G(s) L- {F(s)G(s)} = (f * g)(t) 1. write the transform as a product whose terms are easy to find the inverse transform of 2. F(s) will be the first part and G(s) will be the second part. Find the Laplace transform of these functions most likely using the table 3. Use the convolution integral to obtain h(t) 1. Take the Laplace transform of all the terms and plug in the initial conditions 2. Solve for F(s) 3. Prepare for the inverse transform process by rewriting the terms if need be 4. Find the inverse transform and use a convolution interval on the last term

Impulse Function

a function that is zero everywhere but at the origin where it is infinetely high. However, the area of the impulse is finite The derivation of an impulse function is not unique. Important result is that the function has zero width and an area of one The unit impulse function is simply the derivative of the unit step function Properties: 1. If the integral includes the origin (where the impulse lies), the integral is one . If it doesn't include the origin, the integral is zero 2. If the impulse is not at the origin (and b > a) then the integral is one at the integral and zero otherwise 1. We replace the value of "t" in the function f(t) by the value of "t" that makes the argument of the impulse equal to 0.

Fundamental matrix method

a fundamental matrix of a system of n homogeneous linear ODE x'(t)=A(t)x(t) is a matrix-valued function P(t) whose columns are linearly independent solutions of the system. 1. compute the successive powers of A to find A^n 2. Use e^At = E (t^n A^n/n!) and simplify the matrix 3. Solution x(t) = e^At x0

Newtons cooling law

dT/dt = k (T-Tm) how fast an object cools T= Temp of tea at time t Tm= medium temperature (room temperature) k < 0 ( for cooling) T= Temp of surrounding environment O(0) = initial temp O(t) = temp at time t dO/dt = rate of change of temperature dO(t)/dt = -k (O(t) - T) O'(t) + kO(t) = kt (linear equation) u(t) = e^kt (e^kt * O(t))' = kTe^kt e^kt O(t) = C + kt int(e^ks ds) = C + kT e^kt -1/k e^kt O(t) = C + T(e^kt -1) O(t) = Ce^-kt + T(1-e^-kt) O(0) = C O(t) = (O(0) = T)e^-kt + T For Small Room, Big Turkey O(t) = C/k1+k2 - k1/k1+k2 (T0-O1) e^-(k1+k2)t T(t) = C/k1+k2 + k2/k1+k2 (T0-O0)e^-(k1+k2)t lim = C/k1+k2 (equilibrium temperature) 1. Separate the variables (get all the T's on one side and all the t's on the other side 2. Anti-differentiate both sides (separable equation) 3. Leave in Previous form or solve for T

separable equations

dy/dx = f(x,y) is separable if M(x)dx = N(y) dy Two forms 1) dy/dx = a(x)b(y) 2) dy/dx = a(x)/b(y) Solve by integration int( M(x)) dx = int (N(y)) dy 1. Get all the y's on the left hand side of the equation and all of the x's in the right hand side 2. Integrate both sides 3. Plug in any given values for the constant of integration (c) 4. solve for y

Series Solutions

p(x)y"+q(x)y'+r(x)y=0 - Assume that we can write the solution as a power series in the form SIGMAan(x-x0)^n and determine what the an needs to be. We will only be able to do this if the point x=x0 is an ordinary point -> p(0) =/ 0 1.Write down the form of the solution and get its derivatives 2. Plug the derivatives into the equation 3. Shift the series to be in terms of x^n 4. Strip the n=0 term 5. Find the recurrence relation 6. Plug in values of n and multiply to obtain a recurrence relation set at ak which is the explicit form of the solution 7. obtain the solution y(x)=a0+a1x+a2x^2+.....+anx^n 8. Collect the terms that contain the same coefficient, factor the coefficient out and write the result as a new series

Homogeneous linear systems of ODE's

system of first order equations d/dt x = Ax where x(t) is a (column) vector of length n and A is an n x n matrix x'(t) = Ax(t) We will assume that A has constant coefficients There are three cases: 1. real distinct 2. real but not distinct 3. some eigenvalues are complex Real Distinct: 1. convert the problem to matrix form if not already 2. find the eigenvalues + eigenvectors using the above method 3. Arrive at general solution written in the form x(t)= c1e^l1t u1 + c2e^l2t u2 Complex eigenvalues: 1. convert the problem into matrix form 2. find/set up the characteristic polynomial 3. obtain l1. If l1 = a + ib then l2= a - ib 4. If u1 is eigenvector of l1 and u2 is eigenvector of l2 then u2 = u1_ (conjugate) 5. Arrive at general solution

Solution Space

the set of all possible solutions for the combinational optimization problem vectors are linear combinations an span a linear space

Dirac Delta Function

used to deal with forcing functions with a large force (or voltage) There are many definitions, we will use D(t-a) = lim fk(t-a) A function that is zero everywhere except one point and at that point it can be thought of as either undefined or having an "infinite" value - an example of a generalized function or distribution #26 in my table 1. Take the Laplace transform of everything in the differential equation 2. Apply the initial conditions 3. Solve for F(s) 4. Verify any partial fractions and their inverse transforms 5. obtain solution - The derivative of the Heaviside function is the Dirac Delta Function

Undetermined coefficients

y" + p(t)y' + q(t)y = g(t) Note: only works for a small class of g(t) 1. Find the complementary solution to the differential -> homogeneous equation 2. Look at g(t) and compare it to the table to find the guess 3. Take the first and 2nd derivatives and plug into the differential equation to determine A,B,C etc 4. Solve for our coefficient. If we find a value it means we guessed correctly

Non - homogeneous equations

y" + p(t)y' + q(t)y= g(t) 1. Solve the associated homogeneous equation y" + py' + qy = O using the methods we have already learned and find Yo Yo = c1y1 + c2y2 2. Find a particular solution Yp using either undetermined coefficients or variation of parameters 3. General solution is written in the form Yg = Yo + Yp

linear equation

y' + p(t) y = g(t) y(t0) = Yo (initial conditions Multiply by u(t) = e^int( p(t)) dt (the integrating factor (uy)' = ug and integrate uy = int ( u(t) g(t)) dt + C y(t) = Cy' + u' int (ug) dt 1. Determine the integrating factor u(x) = e^$pdx 2. Multiply the linear equation by the integrating factor 3. Integrate both sides of the equation 4. Undo the substitution 1. Rewrite in the standard form 2. Multiply both sides by the integrating factor: u(x)=e^$pdx 3. Left hand side collapses into (uy)' = uq 4. Integrate both sides of the equation 5. Solve for the constant

Bernoulli equation

y' + p(t) y = q(t) y^m when m >_ 2 the equation is nonlinear Substitute v = y^1-m to obtain a linear equation and then we integrate the now linear equation in v 1. Transform the equation by multiplying through by y^-n y^-n y1 + p(x)y^1-n = q(x) 2. Use substitution w = y^1-n w' = (1-n)y^-n y' 1/1-n w' + p(x)w = q(x) 3. Solve the now linear equation