Ejercicios 7.3 Problema 21 Dennis G. ZILL ED 9na Ed. Transformada de Laplace ED PVI
How to Solve an Initial Value Problem Using Laplace Transform
Introduction to the Problem
- The video introduces the topic of solving a differential equation with an initial value problem using the Laplace transform.
- It emphasizes the need to apply the Laplace transform to convert the differential equation into an algebraic equation.
Steps in Applying Laplace Transform
- The speaker explains that they will separate terms and apply the Laplace transform, focusing on constants as factors similar to integrals and derivatives.
- They mention obtaining transforms for specific functions, particularly noting how to handle derivatives during this process.
Deriving Transforms
- The first derivative is discussed, where it’s noted that s cdot Y(s) - y(0) is used for transformation.
- The speaker calculates the transform of an exponential function, emphasizing its formula and substituting values correctly.
Substituting Transforms into Equation
- After obtaining necessary transforms, they substitute these back into their original equation involving Y(s) .
- A rearrangement of terms leads them towards isolating Y(s) , preparing for further simplification.
Finalizing Expression and Inverse Transform
- The expression for Y(s) is simplified down to a fraction which can be factored further.
- They break down Y(s) , separating it into two fractions for easier handling during inverse transformation.
Application of Inverse Transform Techniques
- The speaker discusses applying inverse transformations using known formulas, specifically addressing exponential functions.
- They highlight using translation theorem techniques when evaluating inverse transforms related to shifts in variables.
Inverse Laplace Transform and Exponential Functions
Understanding the Inverse Laplace Transform
- The discussion begins with the evaluation of the inverse Laplace transform, specifically focusing on a function evaluated at s - at . The speaker emphasizes performing a direct inverse transformation.
- The inverse transform of 1/s-a is noted to be e^at , where in this case, a = -4 . Thus, it results in e^-4t .
- The speaker references previous chapters or sections regarding the inverse transform of 1/s^2 , which is simply represented as the variable multiplied by time ( t ).
- It is clarified that the result from the inverse transform of 1/s^2 will yield an expression involving both time and an exponential factor due to multiplication with e^-4t .
- A comparison is made with a textbook solution for problem 21 from exercise group 7.3, confirming that the derived answer matches expectations: it includes terms like e^-4t + 2e^-400 , indicating rearrangement rather than alteration of values.
Conclusion and Engagement