Consider the ODE ƒ(x, y) = x – y. Take initial conditions x, = 0, yo = y(0) = 11, x1 = 0.2 and y1 = y(0.2) = 2 solve for y(30) with a step size of 0.2 using an adaptation of Euler's method, which uses two known y solutions approximate the next y solution. The general formula is: Yi+2 = Yi+1 +h An example of the first few iterations is shown below. y2 = yı = 16.5600 y3 = Y2 + hf(x2, y½) (x1, Yı) = 13.8920 (3 = 11.5204

For the following pairs of functions, first decide whether f(n) dominates g(n), or g(n) dominates f(n); then decide whether f = O(g), or f = Ω(g), or f = Θ(g), and briefly explains. i. f(n)=n2,g(n)=1000n+30 √1 ii. f(n) = n, g(n) = n3 iii. f(n) = 10 log2 n, g(n) = log10(n3) iv. f(n)=n 100 n ,g(n)=1.2 . 4 nn v. f(n)=2 ,g(n)=2.01 .

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?