Consider the IVPs:
(A) y'+2y = 1, 0<t<1 , y(0)=2.
(B) y' = y(1-y), 0<t<1 , y(0)=1/2.
1. For each one, do the following:
a. Find the exact solution y(t) and evaluate it at t=1.
b. Apply Euler's method with Δt=1/4 to find Y4 ≈ y(1).
Make a table of tn, Yn for n=0,1,2,3,4.
c. Find the error at t=1.
2. Euler's method is obtained by approximating y'(tn) by a forward finite difference.
Use the backward difference approximation to y'(tn+1)...
Let Q1=y(1.1), Q2=y(1.2), Q3=y(1.3), where y=y(x) solves...
1) y'''+2y''-5y'- 6y=4x^2 where y(0)=1, y'(0)=2, y''(0)=3
2) y'''- 6y''+11y'- 6y=6e^(4x) where y(0)=4, y'(0)=10,
y''(0)=30
3) y''- 6y'+9y=4e^(3x) ln(x) where y(1)=, y'(1)=2
Please show all steps and thank you!!!
Find the equation of the ellipse with foci at (0, 0) and (2, 2),
with eccentricity e = 0.5. Express the equation in standard form
ax2 + by2 + cxy + dx + ey = f and in terms of the distance formula
sqrt(x^2+y^2) + sqrt[(x-2)^2 +(y-2)^2]=?
There is an answer posted on Chegg, but I don't think I agree
with it. Since the foci are at (0, 0) and (2, 2) it seems that the
major axis is rotated...
How to transform x^2+xy+y^2+4x+2y=0 into the standard for of an
ellipse and finding the vertices of both major and minor axis. Plot
points and graph the ellipse.
Consider the parabolas y=x^2 and y=a(x-b)^2+c, where a,b,c are
all real numbers
(a) Derive an equation for a line tangent to both of these
parabolas (show all steps with a proof, assuming that such
a line exists)
(b) Assume that the doubly-tangent line has an equation y+Ax+B.
Find an example of values of a,b,c (other than the ones given here)
such that A,B ∈ Z