I'm having trouble understanding what exactly makes a linear ordinary differential equation "linear". I know that the major advantage of the linear ODEs is that a linear combination of particular solutions gives another particular solution. Characterized by additivity and homogeneity, this means that the output for a sum of inputs is equal to the sum of outputs for each individual input (e.g., f(3) + f(5) = f(3+5)= f(8)) and scaling the input by a factor scales the output by the same factor (e.g., f(6x) =6f(x)...but how would this look? Like say if I had 4e^5(t-1) satisfying the differential equation dy/dt=5y where y(1)=4, what would that "linear combination" look like?

. Obtain the canonical SOP form of the following functions:                                    (i) y(A,B) = A+B (ii) f (A,B,C) = A+BC

Show me the steps of determine red

Develop a Matlab function that finds a root of a function f(x) starting from the given initial estimates (-¹) and (0) with a tolerance in function of at least OK using the secant method. Name the function mySecant using as input the anonymous function f, the initial estimates xm1 and x0, the maximum number of iterations to perform N, and the required tolerance in function epsok. As output, the function shall return four scalar variables: the numerical solution x, its tolerance in function tolFunc, its estimated relative error ere, and the number of iterations performed n. If the requested tolerance in function cannot be reached within N iterations, the function shall execute Matlab's error ("...") function with an appropriate error message. Other than this potential error message, do not print out any results to screen or do any plotting within the function. You must minimize the number of function calls required per iteration, by storing function call results in temporary storage…

This is discrete math:

Can you represent the following Boolean function f(A, B) with a linear classifier? If yes, show the linear classifier. If not, briefly explain why not. f(A, B) A B 1 1 1 1

Write Matlab program to plot the functions: f1 = sin(2x)  f2 = 2sin(x)cos(x) Given x = 0:0.01:2pi

2. calculates the trajectory r(t) and stores the coordinates for time steps At as a nested list trajectory that contains [[xe, ye, ze], [x1, y1, z1], [x2, y2, z2], ...]. Start from time t = 0 and use a time step At = 0.01; the last data point in the trajectory should be the time when the oscillator "hits the ground", i.e., when z(t) ≤ 0; 3. stores the time for hitting the ground (i.e., the first time t when z(t) ≤ 0) in the variable t_contact and the corresponding positions in the variables x_contact, y_contact, and z_contact. Print t_contact = 1.430 X_contact = 0.755 y contact = -0.380 z_contact = (Output floating point numbers with 3 decimals using format (), e.g., "t_contact = {:.3f}" .format(t_contact).) The partial example output above is for ze = 10. 4. calculates the average x- and y-coordinates 1 y = Yi N where the x, y, are the x(t), y(t) in the trajectory and N is the number of data points that you calculated. Store the result as a list in the variable center = [x_avg, y_avg]…