Recent content by ch2kb0x

[bump] Question revised: 1) Apply Runge-Kutta of order 4 to solve the ODE on [-1, 1] with N=4. At the very least solve for the points in the interval that are in [0,1]. Is it possible with the initial condition y(0)=1 to obtain numerical solutions in [-1,0) using Runge-Kutta? If so then how can...

Homework Statement e^(x) + y = dy/dx, [-1,1], y(0) = 1, N = 4. Homework Equations The Attempt at a Solution h = b-a / N = 0.5 x0= 0, y0 =1 x1= 0.5, y1 = 2.472 x2= 1, y2 = 5.433 x3= 1.5, y3= 11.195 x4=2, y4= 22.146 ===================== These were the values I got for the...

Yeah, it is a function of x and y, but I don't think it's in the g(y/x) form.

Yeah, copied exactly from textbook.

Okay, so since you said y = ux, I am thinking that this is a homogenous equation... However, if it is a homogeneous equation, before we can plug in y = ux, aren't we suppose to first have the equation in the form of dy/dx = f(x,y), where there exists a function such that f(x,y) is expressed...

Homework Statement (x + 2y) dy/dx = 1, y(0) = 1 Homework Equations The Attempt at a Solution Problem is, I can't separate it. This might be a homogenous type? If so, how would I make it into the g(y/x) form. Thank you.

Hmmm ok, still kind of confused. this is what I did after your correction: dx/dy + x = 2e^y I(y) = e^integral (2e^y) dy = e^(2e^y) multiply integrating factor on both sides will give: e^(2e^y) x = Integral e^(2e^y) (2e^y) dy...i am unsure on how to solve that integral on the right side.

So, if I were to solve, would it be like this: dx/dy + x = 2e^y I(x) = e^integral dy = e^y multiply integrating factor on both sides will give: e^y (x) = e^(2y) + C => x = e^y + Ce^-y. That was the answer I got for x, but it does not match answer in the back. help.

Homework Statement (2e^y -x) dy/dx = 1 Homework Equations dy/dx + P(x)y = Q(x) The Attempt at a Solution I know how to solve these equations, but I can't get this into the dy/dx + P(x)y = Q(x) form. In addition to this problem, is also this: (x + y^2)dy = ydx (can't get into...

For dy/dx - y = 4e^x, with y(0) = 4. I got answer: y = (4x + C)e^x The answer in back says 4e^x (x + 1) What am I doing wrong? My Procedure: P(x) = -1 Q(x) = 4e^x I = e^-x Integral both sides = > e^-x(dy/dx -y) = e^(-x) (4e^x)dx = e^-x y = 4x + c = y = (4x + c )e^x

Nm, i know how to solve it now, thanks. Although, I am stuck on this problem now: dy/dx - y = 4e^x, with y(0) = 4. How do I solve this when there isn't a P(x) function next to y...or is there...

Homework Statement dy/dx + ytanx = secx, y(pi) = 1 Homework Equations I(x) = e^integral P(x)dx Integral Q(x)e^integral P(x)dx y=e^-integral P(x)dx (integral Q(x)e^integral P(x)dx + C) (sorry, it is a bit messy, I don tknow how to use the math symbols yet) The Attempt at a...

Question, I am doing the same problem, but i do not understand why a (-) sign is put for part A) but not B) & C).

Homework Statement A projectile fired down the y-axis with V0 = 300 ft/s, hits a target 500ft away. Assume constant force is acting on projectile at 60ft/s in negative z direction. Assume second constant force acting on projectile in negative x direction at 45ft/s. Find angle of elevation...

Homework Statement Water flowing out of a 16-mm diameter faucet fills a 2L bottle in 10s. At what distance below the faucet has the water stream narrowed to 10mm diameter? Homework Equations A1v1=A2v2 Q=vA The Attempt at a Solution edit: this is what I TRIED to do:(*note: i ignored some...