a 3rd order IVP I am havin trouble with:
y''' -3y'' +2y' = t + e^t y(0)=1, y'(0)= -.25 y''(0)= -1.5
I am using At^2 and B*e^t *t as my Y1 and Y2. Is this correct?
I am stuck on solving for the roots of a charactristic equation:
y'''- y''+y'-y=0
where I set r^3-r^2+r-1=0 and factored out r to get r*[ r^2-r +1] -1 =0 to get the real root of 1. How can I solve for the compex roots?
Given t^2 y'' -t(t+2)y' = (t+2)y= 2t^3 and y1= t, y2= te^t
Find the particular solution-
I ve worked the problem to [ -2t^2 -2t] by:
-t * Integral [ 2t* te^t/ t^2e^t] + te^t * Integral [ 2t^2/ t^2e^t]
whereas the book states that it is simply -2t^2. Can you guys tell me where I made...
The problem is :
(y*e^(2xy) +x) + [ b* x*e^(2xy) ] y' =0. Find b so the equation is exact and solve.
I found b=1 and worked the problem to (1/2)e^(2xy) + (x^2/2)e^(2xy) + h(y); where I found h(y) to be simply c.
The answer in the text states that e^(2xy) + x^2 =c
Would I be...
I am having trouble with the below:
[ 4* (x^3/y^2) + (3/y)] dx + [3*(x/y^2) +4y]dy=0
I found My= -8x^3y^-3 - 3y^-2 and Nx= 3y^-2
i then subtracted Nx from My and divided by [3*(x/y^2) +4y]
[-8x^3y^-3 - 6y^-2] / [3*(x/y^2) +4y]. can you guys give me a hint as to where my error is?
Can you guys point me in the right direction on the problem below?
Solve the given initial value problem and determine at least approx. where the solution is valid:
(2x-y)dx + (2y-x)dy= o, y(1)=3
So I have My =-1 and Nx= -1
x^2-xy+ h(y) => -x+h'(y) = 2y-x => h(y)= y^2
=> x^2...
I am having trouble with part of this question:
y'+2y= g(t) where g(t) =1 for 0<t<1 and g(t)= 0 t>1 and y(0)=0
I found y= 1/2 - (1/2 * e^-2t) for 0<t<1 but am having trouble with g(t)= 0.
I know y = Ce^-2t. y(0)=0 doesn't apply here since t>1 correct? Can you point me in the right...