Recent content by EnzoF61

Very insightful. Thank you halls!

x' = y-2x y' = 2x-y Whenever you look for equilibrium points x'=0 and y'=0, y=2x for both cases. What is the meaning of this outcome? Infinite equilibrium points?

If \overline{_}X1 and \overline{_}X2 are the means of independent random samples of size n1 and n2 from a normal population with the mean \mu and \sigma^2, show that the variance of the unbiased estimator Var(\omega\overline{_}X1 +(1-\omega)\overline{_}X2) is a minimum when \omega= n1 / (n1 +...

Are you interpreting a_3 as another coefficient in my work? I.e. a_1*a_2 = a_3

I'm coming closer, but my results are still not precise. y = a_0 + a_1x + a_2x^2 y(0) = 0 y(0) = a_0 + 0 + 0 = 0 ==> a_0 = 0 y = a_1x + a_2x^2 y' = a_1 + 2a_2x (y)^2 = (a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4 (y')^2 = (a_1 + 2a_2x)^2 (y')^2 = 1 - (y)^2 (a_1 + 2a_2x)^2...

The homework has already been turned in, but I challenge you to prove me wrong and solve this numerically. http://www.math.pitt.edu/~troy/classes/math1270.html You can see that Homework 3 was due on the 18th The book was Boyce and DiPrima 9th Ed. Elementary Diff Eq and Boundary Value...

I have still been working with this problem. I've tried general series, squaring them, binomial expansions. Equating coefficients. This is IMPOSSIBLE to solve numerically.

I don't think that it is possible start with a series solution. The question says look for a solution of the initial value problem in the form of a power series. Not only can we can show that this is a solution of the initial value problem, but by using Taylor's expansion, sin(x) is in the form...

I believe that it is impossible to obtain a recursion series solution up to the x^3 term. However I did recognize that y' = sqrt(1-y^2) dy/dx = sqrt(1-y^2) dy/sqrt(1-y^2) = dx integral (dy/sqrt(1-y^2)) = integral (dx) inverse sin (y) = x y = sin(x) y(0) = 0 y =...

y = n=0--> ∞ sum(a_n*x^n) y' = n=1-->∞ sum(n*a_n*x^n-1) y' = sqrt(1-y^2) y = sqrt(1-(y')^2) y(0) = 0 sqrt(1-(y')^2) = 0 (n=1 --> ∞ sum(n*a_n*x^n-1))^2 = 1 Solved for y = 0 Now I have sum for (y')^2 = 1 I am having ambiguity with the general series for y(x) = a_0 +...

Hmm Let y = sinx be a solution of y' = sqrt(1-y^2) integral ( y') = integral {sqrt( 1 - sin^2(x) )} integral (dy) = intergral {cos(x) dx} y = sin(x) Given y(0) = 0 We have that, y(0) = sin(0) = 0 ==> y = sin(x) is a solution of the initial value problem Hence y =...

Homework Statement y' = \sqrt{(1-y^2) } Initial condition y(0) = 0 a) Show y = sinx is a solution of the initial value problem. b) Look for a solution of the initial value problem in the form of a power series about x = 0. Find coefficients up to the term in x^3 in this series...

Homework Statement y' = \sqrt{(1-y^2) } Initial condition y(0) = 0 a) Show y = sinx is a solution of the initial value problem. b) Look for a solution of the initial value problem in the form of a power series about x = 0. Find coefficients up to the term in x^3 in this series. Homework...