Recent content by JamesEllison

That is truly amazing :D three cheers for Maths!

Ah. Terriffic. y = ax-3 + bx\frac{5}{2} Apply IC1: y(1) = 0 y(1) = a(1)-3 + b(1)\frac{5}{2} = 0 y(1) = a + b => a = -b Then \frac{dy}{dx} = -3ax-3 + \frac{5}{2}bx\frac{5}{2} IC2: y'(1) = 1 1 = -3a + \frac{5}{2}b Since a = -b 3b + \frac{5}{2}b = 1 \frac{6}{2}b...

On phone again xD "Put x =e^t ; solve for Y against t." y = Ae^mx + Be^mx y = Ae^me^t + Be^me^t? Just before I go ahead.. http://tutorial.math.lamar.edu/classes/DE/eulerequations.aspx

Thanks very much. Officially on a computer now. My screen was far to small, and i was writing out by hand first to see what they'd really look like. :) Thanks for being patient. 2\frac{d^2y}{dt^2} - 2\frac{dy}{dt} +3\frac{dy}{dt} - 15y = 0 2\frac{d^2y}{dt^2} +\frac{dy}{dt} - 15y = 0...

Cool, So the assumption of my quadratic formula should be: 2m2 + 3m - 15 = 0 m = (-3±√129)/4 m = 2.089 m = -3.589 So that general assumption is: Aemx + Bemx = y Ae2.089x + Be-3.589x = y IC 1 y = Ae2.089 + Be-3.589 = 0 IC2 dy/dx = 2.089Ae2.089 - 3.589Be-3.589 = 1 How...

Excellent. Sorry for the lack of real looking equations, I am on my phone. 2m^2 + 3m - 1 = 0 m = [-3+/- sqrt( 9 + 8 )]/4 = 1.1231 / 4 =0.2808 m = -7.1231/4 = -1.7808 Should y be assigned a coefficient? Ae^mx + Be^mx = Cy Ae^0.2808x + Be^-1.7808x = Cy Sub in IC 1 x = 1 y = 0...

Problem: Solve the initial Value: when x=1, y=0 dy/dx = 1 2x^2(d^2y/dx^2) + 3x (dy/dx) - 15y = 0 My attempt: x = e^t dx/dt = e ^t dy/dt = dy/dx * dx/dt dy/dt = x*dy/dx d^2y/dt^2 = d/dt(dy/dt) = d/dt(x*dy/dx) =d/dx(x*dy/dx)*(dx/dt) since dx/dt = x =(x^2*d^2y/dx^2) + (x*dy/dx)...

Excellent. So from there: Yp = C Sin(2x) Yp' = 2C Cos(2x) Yp" = -4C Sin(2x) -4C Sin(2x) + C Sin(2x) = Sin(2x) -3C = 1 C = -1/3 Yp = -1/3 Sin (2x) So then the solution starts as Y = Yg + Yp Y = A Sin (x) + B Cos(x) -1/3 Sin(2x) Sub in initial Conditions Q(0) = 1 or x=0 Y = A Sin (x) + B...

Or just Yg = A Sin2x + i BCos 2x ? Sorry about lack of real looking equations I am on my iPhone

e^kt(k^2+1) = 0 k = i ---> sqrt(-1) So then the general assumption would be. Yg = e^-kt(ASin2x + iBsin2x) ?

I see. Thanks.

Homework Statement Solve the ODE with the boundary conditions given Q''+Q = Sin(2x) where Q(0) = 1 and Q'(0) = 2 So i know i need to solve the general and particular solutions, however, I am a little confused. Any help or advice would be great, Thanks in advance. Homework Equations Y...