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.
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.
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
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.
The solution to the Heat equation using Fourier transform is given by the convolution of the initial condition with the fundamental solution of the heat equation, which is the Gaussian function. The Fourier transform helps in solving the heat equation by transforming the problem from the spatial domain to the frequency domain, simplifying the calculations.
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
Newton's method, also known as Newton-Raphson method, is an iterative technique for finding the roots of a real-valued function. It starts with an initial guess and refines the estimate in each iteration by using the derivative of the function. The method is based on the principle that a function can be approximated locally by a linear function at a root.
Two
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].
The word "initial" can function as either an adjective or a noun.