Just A Simple Fouier Transform Example:

Given x(t) = exp(-2*t) u(t) and X(ω) is equal to 1/(j*omega + 2);

Use The Frequency Differentiation of Fourier Transform to the Given Problem and Plot With An Amplitude Spectrum Graph:

Sorry Everyone: I am trying To Learn This Software and Need A Bit Of Help: I am Trying To Solve & Plot With An Amplitude Spectrum Graph #A & Problem #B. Can Anyone Help: 

Sorry I Do Not Know How & I Can Find Very Little Via Google a Google Search and Thank You A Million Times For Anyone That Can Help Me With Problems. 

#A). t*x(t)
#B). t^2*X(t)

#Solution To Example Problem #A 't*x(t)':

(t)^nX(t)->(j)^n*((ⅆ)^n)/((ⅆ)^( )omega^n)(X(omega));

n := -1;
                        
X := omega -> 1/(2 + j*omega);

diff([j^n*diff(X(omega), [omega $ n])], omega);

Hi Everyone, 

I am searching for some examples on Maple how to do a Static Analysis of a Truss: Method of Joints. IS there any Maplesoft App or something that has been built already that anyone here can recommend to me. Thank you in advance.  

Hi everyone, I need someone’s expertise with the Maple Soft Application, I am currently into my 4^th day of my Maple Soft experience and also returning to School after many years. Can someone help me with the following codes to solve step by step Differential Equation 1 below;

Please help me with following codes below

1. Boundary Condition 1
2. Boundary Condition 2
3. General Solution
4. The integration Part of the Equation
5. Solving the Differential Equation  

Equation 1.

Determine the equilibrium temperature for a one dimensional rod with a constant thermal properties with the following source and boundary conditions.

1. Q = 0, U(0) = 10, u(L) = 20
2. Q/K0 = x, u(0) = 0, u(L) = 10

I understand how to obtain the solution to the equation above,

(a) Equilibrium satisfies

                                                U’’(x) = 0,

Whose general solution is u = c1 + c2x.

The boundary condition u(0) = 0 implies c1 = 0 and u(L) = T implies c2 = T/L so that u = T x/L.

(f) In equilibrium, u satisfies

U’’(x) = −Q/K0 = −x^2,

Whose general solution (by integrating twice) is

u = −x ^4 /12 + c1 + c2x.

The boundary condition u(0) = T yields c1 = T, while u’(L) = 0 yields c2 = L^3/3.

Thus u = −x ^4 /12 + L^3x/3 + T.