Solve y''+y=0 using Laplace.
Umm y=0, 0''+0=0, 0.o Oh well here it is.
First you take the Laplace of each term, so . . .
L(y'')+L(y)=L(0)
Using your Laplace table you know the Laplace of all these terms
s2L(y)-sy(0)-y'(0) + L(y) = 0
Since both initial conditions are 0 this simplifies to. . .
s2L(y) + L(y) = 0
You can factor out the L(y) and solve for it.
L(y) = 0/(s2+1)
L(y) = 0
Now take the inverse Laplace of both sides and solve for y.
L-1(L(y)) = L-1(0)
y = 0
All of the following are responsibilities of derivative classifiers EXCEPT: Derivative classifiers must have access to classification guidance. Derivative classifiers must understand derivative classification policies and procedures. Derivative classifiers must have original classification authority. Derivative classifiers must possess the requisite subject matter expertise, as well as classified management and marking techniques.
All of the following are steps in derivative classification EXCEPT: Seek additional guidance to resolve uncertainty Analyze material to be classified Use authorized sources for guidance Make recommendations for others to mark the new document
Let us say that f(x)=x^4A derivative is the opposite to an integral.If you were to integrate x^4, the first process is taking the power [which in this case is 4], multiplying it by any value before the x [which is 1], then subtracting 1 from the initial power [4]. This leaves 4x^3. The final step is taking the integral of what is 'inside' the power [which is (x)], and multiplying this to the entire answer, which results in 4x^3 x 1 = 4x^3If you were to derive (x)^4, you would just add 1 to the power [4] to become (x)^5 then put the value of the power as the denominator and the function as a numerator. This leaves [(x^5)/(5)]To assure that the derivative is correct, integrate it. (x^5) would become 5x^4. Since (x^5) is over (5), [(5x^4)/(5)] cancels the 5 on the numerator and denominator, thus leaving the original function of x^4
If you want to find the initial value of an exponential, which point would you find on the graph?
using Laplace transform, we have: sY(s) = Y(s) + 1/(s2) ---> (s-1)Y(s) = 1/(s2), and Y(s) = 1/[(s2)(s-1)]From the Laplace table, this is ex - x -1, which satisfies the original differential eq.derivative of [ex - x -1] = ex -1; so, ex - 1 = ex - x - 1 + xto account for initial conditions, we need to multiply the ex term by a constant CSo y = C*ex - x - 1, and y' = C*ex - 1, with the constant C, to be determined from the initial conditions.
the velocity function v= at + v(initial)
I suggest: - Take the derivative of the function - Find its initial value, which could be done with the initial value theorem That value is the slope of the original function.
Laplace will only generate an exact answer if initial conditions are provided
to incorporate initial conditions when solving difference equations using the z-transform approach
If you are only given total distance and total time you cannot. If you are given distance as a function of time, then the first derivative of distance with respect to time, ds/dt, gives the velocity. Evaluate this function at t = 0 for initial velocity. The second derivative, d2s/dt2 gives the acceleration as a function of time.
If you are actually in your car, check the spedometer. That will tell you your instantaneous velocity; that is, distance traveled per second.If this is a calculus question and you are given the function of your position with respect to time, simply take the derivative of your function and evaluate your derivative at the time at which you would like to determine your instantaneous velocity.Alternatively and more unlikely, you can integrate your acceleration function and solve for your antiderivative based on an initial value given by the context of the problem.
applicable only to LTI s/m no account of initial conditions considered no idea abt present i/p
A transfer function (also known as the system function[1] or network function and, when plotted as a graph, transfer curve) is a mathematical representation, in terms of spatial or temporal frequency, of the relation between the input and output of a linear time-invariant system with zero initial conditions and zero-point equilibrium. With optical imaging devices, for example, it is the Fourier transform of the point spread function (hence a function of spatial frequency) i.e. the intensity distribution caused by a point object in the field of view. An alternative brief definition is "a mathematical function relating the output or response of a system such as a filter circuit to the input or stimulus"[2].
Boundary conditions allow to determine constants involved in the equation. They are basically the same thing as initial conditions in Newton's mechanics (actually they are initial conditions).
Two
Integration is the opposite of differentiation (taking the derivative). The derivative of a constant is zero. Integration is also called antidifferentiation since integration and differentiation are opposites of each other. The derivative of x^2 is 2x. The antiderivative (integral) of 2x is x^2. However, the derivative of x^2 + 7 is also 2x. Therefore, the antiderivative of 2x is x^2 + C, in general, where the constant C has to be determined from the context of the problem. In the above case, the constant happens to be C=7. We use integration to solve first order differential equations. When solving first order differential equations, like in "word problems", you must determine the integration constant using the initial conditions (ie the conditions we know to be true at t=0 - we usually know what these are), or the boundary conditions (ie the conditions we know to be true at x=0 and y=0).
Initial Conditions - 2007 was released on: USA: 15 April 2007 (First Run Film Festival)