The solution to the Initial value problem
x′′+2x′+17x=2cos(6t),x(0)=0,x′(0)=0 is the sum of the steady
periodic solution...
The solution to the Initial value problem
x′′+2x′+17x=2cos(6t),x(0)=0,x′(0)=0 is the sum of the steady
periodic solution xsp and the transient solution xtr. Find both xsp
and xtr.
Solve the initial value problem
y′=(2cos(2x))/(3+2y), y(0)=−1
and determine where the solution attains its maximum value (for
0≤x≤1.697).
Enclose arguments of functions in parentheses. For example,
sin(2x).
Y(x)=?
x=?
Homework 1.1. (a) Find the solution of the initial value problem
x' = x^(3/8) , x(0)=1 , for all t, where x = x(t). (b) Find the
numerical solution on the interval 0 ≤ t ≤ 1 in steps of h = 0.05
and compare its graph with that of the exact solution. You can do
this in Excel and turn in a printout of the spreadsheet and
graphs.
Solve the given initial-value problem. y'' + 4y' + 4y = (5 +
x)e^(−2x) y(0) = 3, y'(0) = 6
Arrived at answer
y(x)=3e^{-2x}+12xe^{-2x}+(15/2}x^2e^{-2x}+(5/6)x^3e^{-2x) by using
variation of parameters but it was incorrect.
1) Find f(x) by solving the initial value problem. f'
(x) = e x − 2x; f (0) = 2
2) A rectangular box is to have a square base and a
volume of 20f t^3 . If the material for the base costs 30¢/f t^2 ,
the material for the sides costs 10¢/ft^2 , and the material for
the top costs 20¢/f t^2 , determine the dimensions of the box that
can be constructed at minimum cost.
Solve the initial value problem dy/dx = −(2x cos(x^2))y +
6(x^2)e^(− sin(x^2)) , y(0) = −5
Solve the initial value problem dy/dt = (6t^5/(1 + t^6))y + 7(1
+ t^6)^2 , y(1) = 8.
Find the general solution of dy/dt = (2/t)*y + 3t^2* cos3t