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##### Solutions to ODEs

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**A general form for a first order ODE is**• Or alternatively**A general form for a first order ODE is**• Or alternatively • We desire a solution y(x) which satisfies (1) and one specified boundary condition.**To do this we divide the interval in the independent**variable x for the interval [a,b] into subintervals or steps.**To do this we divide the interval in the independent**variable x for the interval [a,b] into subintervals or steps. • Then the value of the true solution is approximated at n+1 evenly spaced values of x. • Such that, • where,**We denote the approximation at the base pts by so that**• The true derivation dy/dx at the base points is approximated by or just • where**We denote the approximation at the base pts by so that**• The true derivation dy/dx at the base points is approximated by or just • where • Assuming no roundoff error the difference in the calculated and true value is the truncation error,**The only errors which will be examined are those inherent to**the numerical methods.**Numerical algorithms for solving 1st order odes for an**initial condition are based on one of two approaches: • Direct or indirect use of the Taylor series expansion for the solution function • Use of open or closed integration formulas.**The various procedures are classified into:**• One-step: calculation of given the differential equation and • Multi-step: in addition the previous information they require values of x and y outside of the interval under consideration**The first method considered is the Taylor series method.**• It forms the basis for some of the other methods.**For this method we express the solution about some**starting point using a Taylor expansion.**Where**• And • etc.**Consider the example, where**• And initial condition**Consider the example, where**• And initial condition • Differentiating and then applying initial conditions,**Consider the example, where**• And initial condition • Differentiating and then applying initial conditions,**Consider the example, where**• And initial condition • Differentiating and then applying initial conditions,**Consider the example, where**• And initial condition • Differentiating and then applying initial conditions,**Continuing,**. . . . Here the third derivative and higher are zero.**Substituting these values into Taylor expansion gives,**• To determine the error lets consider compare this to the analytic solution.**Substituting these values into Taylor expansion gives,**• To determine the error lets consider compare this to the analytic solution. • To do this we will integrate the differential equation.**Integrating,**• Simplifying,**In this case the two expressions are identical, therefore**there is no truncation error.**Homework (Due Thursday)**• Consider the function • Initial condition • Write the Taylor expansion and show that it can be written in the form, • Show if terms up to are retained, the error is,**Stepping from to follows from the Taylor**expansion of about .**Stepping from to follows from the Taylor**expansion of about . • Equivalently stepping from to can be accomplished from the Taylor expansion of about .**Stepping from to follows from the Taylor**expansion of about . • Equivalently stepping from to can be accomplished from the Taylor expansion of about .**Algorithms for which the last term in expansion is dropped**are of order hn. • The error is or order hn+1. • The local truncation error is bounded as follows • where**However differentiation of can be complicated.**• Direct Taylor expansion is not used other than the simplest case, • Big Oh is the order of the algorithm.**Usually only the value of is the only value of**that is known, therefore must be replaced by . • Thus, • In general, which is known as Euler’s method.**The problem with the simple Euler method is the inherent**inaccuracy in the formula.**y**y1 y(x) y(x0) y(x1 h x0 x1 x • The geometric interpretation is shown the diagram.**y**y1 y(x) y(x0) y(x1 h x0 x1 x • The solution across the interval[x0,x1] is assumed to follow the line tangent to y(x) at x0.**When the method is applied repeatedly across several**intervals in sequence, the numerical solution traces out a polygon segment with sides of slope fi, i=0,1,2,…,(n-1).**The simple Euler method is a linear approximation, and only**works if the function is linear (or at least linear in the interval).**The simple Euler method is a linear approximation, and only**works if the function is linear (or at least linear in the interval). • This is inherently inaccurate. • Because of this inaccuracy small step sizes are required when using the algorithm.**From the graph, as h -> 0, the approximation becomes**better because the curve in the region becomes approx. linear. y y1 y(x) y(x0) y(x1 h x0 x1 x**The modified Euler method improves on the simple Euler**method.