Use the Laplace transform to solve the given initial value problem. y^''+ω^2 y=cos2 t, ω^2 ≠4

Use the Laplace transform to solve the given initial value...

Key Concepts

Inverse Laplace Transform

After solving for the function in the Laplace domain, the inverse Laplace transform is used to convert the solution back to the time domain. This process often involves techniques such as partial fraction decomposition to simplify the expression into known transform pairs, enabling the recovery of the original solution.

Initial Conditions in Laplace Transform Method

The Laplace transform method naturally incorporates initial conditions when transforming derivatives. This means that the initial values of the function and its derivatives are directly used in the algebraic equation in the Laplace domain, ensuring that the final solution satisfies these initial conditions without the need for additional steps.

Transformation of Derivatives

A key property of the Laplace transform is its handling of derivatives. For instance, the Laplace transform of a first derivative involves both the transform of the function and its initial value. This property simplifies the incorporation of initial conditions directly into the transformed equation and is essential for solving initial value problems.

Solving Differential Equations with Laplace Transforms

Using the Laplace transform to solve differential equations involves taking the transform of both sides of an ODE. This leverages the linearity of the transform and the properties that simplify the handling of derivatives, ultimately reducing the problem to solving an algebraic equation in the Laplace domain before inverting the result to recover the solution in the original domain.

Laplace Transform

The Laplace transform is an integral transform that converts a function of time into a function of a complex variable, usually denoted by s. It is particularly useful for solving linear ordinary differential equations (ODEs) because it transforms differential equations into algebraic equations that are easier to manipulate and solve.

Transcript

00:05 In this question, we are given a differential equation, which is y double prime plus omega squared y prime equals cost 2d.

00:30 And we are given values, boundary values, y 0 is equal to 1, and y prime 0 is equal to 0.

00:51 So first i will convert this using laplace transform.

00:58 This will be s squared capital ys minus s simple by zero minus y prime zero plus omega squared.

01:27 I made a mistake here.

01:28 This is y omens squared y s is equal to the laplace transform of cost 2d which is s over s squared plus 4 so cost laplace transform or cost a t is equal to s over s squared plus a squared so a is 2 here so it's going to be 4.

02:12 And we can substitute the values, y0, and y prime 0.

02:22 And we can simplify this to s squared ys minus s plus omega squared ys is equal to s over s squared plus s squared plus 4.

02:42 I will further simplify this.

02:47 S squared plus omega square times ys is equal to, i'm going to carry this s to the other side, which gives us s over s squared plus 4 plus s.

03:11 So this is going to give us 1 over s squared plus omega squared times if you solve this, this is going to be sq plus 5s over s squared plus 4.

03:43 Now i'm going to convert this expression to partial fractions like this.

03:55 Sq plus 5s over s squared plus omega squared into s squared plus 4 is equal to as plus b over s squared plus omnisquare plus c s plus d over s squared plus 4 and i'm going to multiply both sides by the common denominator and this will give us as plus b times s square plus 4 plus c s plus d times s squared plus omnisquare.

05:09 And i will equate the coefficients of similar terms.

05:17 So we know that multiplication of this, and this gives the cubic term.

05:27 So 1 is equal to a plus d, a plus c, and then i'm going to equate the constant terms, which gives us 0 is equal to 4b plus d omega squared.

06:42 And then this term into this term gives the first order term, and it gives us 5 is equal to 4a plus c omega squared.

07:16 And first i will consider these two.

07:22 And if i multiply 1 by 4, it's going to give us 4 is equal to 4a plus 4c...

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