# Solving Initial Value Problem

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The comparison of the implemented Maple package with existing methods implemented in other mathematical software like Matlab and Mathematica is also discussed in Results Section.

In this paper, we focused on Maple implementation of an IVP with homogeneous initial conditions, however we also discuss an algorithm to check the consistency of the non-homogeneous initial conditions.

We get $(s^2\mathcal\ - 2s - 1) (s\mathcal\ - 2) - 2\mathcal\ = 4/s.$ Next, combine like terms to get $(s^2 s - 2)\mathcal\ = 4/s 2s 3. e^ f(t) \right|_0^ s\int_0^ e^f(t)\,dt = -f(0) s\mathcal\$ For functions $$f$$ such that $$f$$, $$f'$$, and $$f''$$ satisfy the theorem's conditions we have $\mathcal\ = -f '(0) s\mathcal\ = -f '(0) s[-f(0) s\mathcal$ \] We will be using this often.

We rewrite the result as $\mathcal\ = s^2\mathcal\ - sf(0) - f '(0)$ This process will work for higher order derivatives also.We begin by applying the Laplace transform to both sides.By linearity of the Laplace transform, we have $\mathcal\ \mathcal\ - 2\mathcal\ = \mathcal\ . There is on in the textbook or you can find one online here.If it is piecewise continuous, we can just break the integral into pieces and the proof is similar.We have \[ \mathcal\ = \int_0^ e^\, f'(t)\,dt$ \[ \left.Sample computations are presented to illustrate the Maple package.Applications of DAEs arise naturally in many fields, for example, various dynamic processes, mechanical systems, simulation of electric circuits and chemical reactions subject to invariants etc., and these are often expressed by DAEs, which consist of algebraic equations and differential operations.We have Proof To prove this theorem we just use the definition of the Laplace transform and integration by parts.We will prove the theorem for the case where $$f$$' is continuous.Then generalized inverse for A and B is calculated, and the problem is reduced to solving a system of ODEs. In Matlab, the equation is also converted to system of ODEs by reducing the differential index and then we find the general solution with free parameters.However, in the proposed algorithm, we compute the exact solution directly without free parameters.

## Comments Solving Initial Value Problem

• ###### Elementary Calculus Solve the Initial Value Problem

Solve the Initial Value Problem. -2y = 0 when y = 0. Thus the constant yt = 0 is a particular solution. where C = eB if y 0, and C = -eB if y 0. General solution yt = Ce-2t, where C is any constant C can be 0 from Step 1. Substitute 1 for t and - 5 for y, and solve for C. -5 = Ce-21, -5 = Ce-2, C = -5e2.…

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In an initial value problem, the ODE is solved by starting from an initial state. Using the initial condition, y 0, as well as a period of time over which the answer is to be obtained, t 0, t f, the solution is obtained iteratively. At each step the solver applies a particular algorithm to the results of previous steps.…

• ###### Solving Boundary Value Problems for Ordinary Di erential Equations in.

This is an initial value problem IVP. However, in many applications a solution is determined in a more complicated way. A boundary value problem BVP speci es values or equations for solution components at more than one x. Unlike IVPs, a boundary value problem may not have a solution, or may have a nite number, or may have in nitely many.…

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• ###### Solve the initial value problem -

Let's look at solving this with a slightly different method. The matrix equation is equivalent to the pair of equations $\displaystyle y_1'= 2y_1+ y_2+ e^x$ and $\displaystyle y_2'= -y_1+ 2y_2$. Differentiate the first equation again to get $\displaystyle y_1''= 2y_1'+ y_2'+ e^x$.…