Initial value problem : Differential equations:
dx/dt = x + 2y
dy/dt = 2x + y
Initial conditions:
x(0) = 0
y(0) = 2
a) Find the solution to this initial value problem
(yes, I know, the text says that the solutions are
x(t)= e^3t - e^-t and y(x) = e^3t + e^-t
and but I want you to derive these solutions yourself using one
of the methods we studied in chapter 4) Work this part out on paper
to...
Consider the curve given by the equation y^2 - 2x^2y = 3
a) Find dy/dx.
b) Write an equation for the line tangent to the curve at the
point (1, -1).
c) Find the coordinates of all points on the curve at which the
line tangent to the curve at that point is horizontal. d) Evaluate
d 2y /dx2 at the point (1, -1).