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ORDINARY DIFFERENTIAL EQUATION (ODE) LAPLACE TRANSFORM

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1.1Definition of the Laplace Transform Let be a function defined on. The Laplace transform of is defined to be the function given by the integral The domain of the transform is taken to be all values of s for which the integral exists. The Laplace transform of is denoted both by And the alternate notation. Laplace transform

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Note: It should be understood that the integral that defines the Laplace Transform is an improper integral, defined by Example 1: Simplest Laplace transform Find the Laplace transform of the constant function, where. Solution: From the definition of the Laplace transform, we have

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Example 2: Laplace transform of Exponential Functions Find the Laplace transform of, where k is any constant. Solution: From the definition of the Laplace transform, we have for

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Theorem 1: Laplace transform, a linear transformation Let f and g be functions whose Laplace transform exists on a common domain. The Laplace transform satisfies the two properties of being a linear transformation: where c is an arbitrary constant. Example 3: Find the Laplace transform

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Theorem 2: Laplace transform of Derivative Let f is a continuous function whose derivative is piecewise continuous on, and if both functions are of exponential order, then both and exists for, moreover, 1.2Properties of the Laplace Transform Theorem 3: Laplace transform of Higher Derivatives Let the functions f, f’, f’’…. are continuous on and is piecewise continuous and if these functions are of exponential order, then the Laplace transform of all these functions exist for, moreover,

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Theorem 3: Translation Property If the Laplace transform exists for, then for. Example 4: Translation Property Find Solution: From Table of Laplace Transform we know that Using Translation Property, we have

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Theorem 4: Laplace Transform If is a piecewise continuous function on and of exponential order then for, Example 5: Find Solution: Since we have

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Example 6: Combining the Rules Find the transform Solution: Starting from the inside, we compute the following Laplace transform in succession using rules from Table 2. STEP 1: STEP 2: STEP 3: STEP 4:

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1.3Inverse Laplace Transform PURPOSE To introduce the inverse Laplace transform and show how it can be found by using tables with the help of some algebraic manipulations, such as the partial fraction decomposition. DEFINITION The inverse Laplace transform of a function is the unique continuous Function f that satisfies which is denoted by. In case all functions f that satisfy are discontinuous on select any piecewise continuous function that satisfies this condition to be.

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Common inverse Laplace Transform 1. 5. 2. 6. 7. 4.

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Example 7: Simple Inverses Find the inverse Laplace Transform of the following functions. (see slide 12) (a) Solution: (b) Solution: (c) Solution:

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Theorem 5: Linearity of the Inverse Laplace Transform If two inverse Laplace transform and exists, then (i) (ii) These conditions state that the inverse Laplace Transform is a linear transformation Example 8: Preliminary Algebraic Manipulation Find the inverse transform of Solution:

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RULES FOR PARTIAL FRACTION DECOMPOSITION If P(s) and Q(s) are polynomials in s, and if the degree of P(s) is less than the degree of Q(s), then exists and can be found by first writing as its partial fraction decomposition. = terms of the partial fraction decomposition Linear Factor: For each factor ( ) in the denominator of Q(s), there corresponds a term of the form in the partial fraction decomposition. Power of Linear Factor: For each power of a linear factor in the denominator of Q(s), there corresponds a term of the form in the partial fraction decomposition.

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3. Quadratic Factor: For each factor in the denominator of Q(s), there corresponds a term of the form in the partial fraction decomposition. Power of Quadratic Factor: For each power of a quadratic factor in the denominator of Q(s), there corresponds a term of the form in the partial fraction decomposition.

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Example 9: Two Linear Factors Find Solution: Since the denominator contains two linear factors, we write Hence

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Example 10: One Linear and One Quadratic Factor Find Solution: Since the denominator contains one linear factors and one Quadratic factor, we try the decomposition Hence

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Example 11: Power of Linear Factor Find Solution: Since the denominator contains one linear factors and one Quadratic factor, we try the decomposition

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Example 12: Completing the Square Find Solution: Since the discriminant of the quadratic expression is negative, it cannot be factored into real linear factors. Hence, we complete the square we get Hence,

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PURPOSE To show how the initial-value problem can be solved by transforming the differential equation into an algebraic equation in, which can then be solved by using simple algebra. We then show how the inverse Laplace Transform can be found, obtaining the solution. 1.4Initial- Value Problems

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DEFINITION: STEPS IN Solving Differential Equations STEP 1: Take the Laplace Transform of each of the given differential equation, obtaining an algebraic equation in the transform of the solution. The initial conditions for the problem will be included in the algebraic equations. STEP 2: Solve the algebraic equation obtained in Step 1 for the transform of the unknown solution. STEP 3: Find the inverse transform to find the solution. Transform problem: Solution of the transformed problem: Solution of the original problem: Typical problem: Solve the transformed problem (Direct) solution

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Example 13: First-Order Equation Solve the initial-value problem Solution: Since the differential equation is an identity between two functions of t, their transform are also equal. Hence By the linearity of the Laplace Transform, we have Using the derivative property and substituting the initial condition into the above equation, we find

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Solving for gives Hence the inverse gives:

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Example 14: Use of the Laplace Transform Solve the initial-value problem Solution: Taking the Laplace transform of both sides we obtain Using the derivative property:

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Substituting the initial conditions and and using the more suggestive notation, we obtain Solving for gives

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Taking partial fraction of solve Hence the solution is

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p4=Plot[((1/3)Sin[t]+(2/3)Sin[2t]),{t,0,4*Pi}] by Mathematica

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Example 15: Differential equation with Damping Solve the initial-value problem Solution: Taking the Laplace transform of both sides Solving for give Substituting the initial conditions

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Finding the undetermined coefficients gives Hence we have

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p5=Plot[((1/2)Exp[-3t]-(2)Exp[-2t]+(3/2)Exp[-t]),{t,0,5}]

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