Solving Initial Value Problem

Solving Initial Value Problem-40
The implemented Maple packege is based on the convertion of the given system into a canonical form using the shuffle algorithm which produces another simple equivalent system, and the canonical system can be solved easily.The Maple implementation includes computing the canonical system and the exact solution of a given IVP.

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.

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.

However, we recall a symbolic algorithm to compute the exact solution of a given system of DAEs (See [1] for further details of the algorithm).

In this paper, we discuss the Maple package of the symbolic algorithm that computes the exact solution.

\] Notice that the coefficient in front of \(\mathcal\) is the characteristic equation of the differential equation. Putting under a common denominator, dividing and factoring we get \[ \mathcal\ = \dfrac .\] To find \(y\), we need to take the Inverse Laplace Transform of the right hand side.

Unfortunately, finding a function \(y\) such that the right hand side is the Laplace transform of \(y\) is not an easy task.

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.


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