Use the Laplace transform to solve the given system of differential equations subject to the giv

Use the Laplace transform to solve the given system of...

Key Concepts

Algebraic Manipulation of Transforms

Once the Laplace transform converts the differential equations into algebraic equations, manipulation of these equations is essential. This involves solving for the transformed variables by techniques such as substitution and elimination, after which the inverse Laplace transform is applied to obtain the solution in the time domain.

Coupled Differential Equations

Coupled differential equations are systems where multiple functions and their derivatives are interdependent, often modeling phenomena with interacting components. Methods like the Laplace transform allow handling these interdependencies by converting each equation into an algebraic form, which can then be solved simultaneously.

Initial Value Problems

Initial value problems specify the values of the unknown functions at a starting point, providing necessary conditions to determine a unique solution for differential equations. The Laplace transform is particularly effective for these problems because it naturally incorporates initial conditions into the transformed equations.

Laplace Transform

The Laplace transform is an integral transform widely used to convert differential equations in the time domain into algebraic equations in the complex frequency domain. This change of domain simplifies the process of solving ordinary differential equations, as differentiation becomes multiplication by the complex variable, making analysis, manipulation, and solution more straightforward.

Inverse Laplace Transform

The inverse Laplace transform is the process of converting a function from the complex frequency domain back to the time domain. After solving the algebraic equations obtained from the Laplace transform, the inverse transform is applied to recover the original function or system response in terms of time.

Transcript

00:01 The first step for this problem is going to be taking the flash transform of our system of equations.

00:06 When we do so, we're going to get s times of the plus transform of x1 minus x1 at 0, minus 2 times the plus transform of x2 is all equal to 0.

00:18 And that's the first equation.

00:20 And the second one, we're going to get x times the plosh transform of x2 minus x2 at 0, plus 2 times the plus transform of x1 is equal to 0.

00:32 Gathering x1 and x2 and putting the constants on the right will give us s comes to the flash transform of x1 minus 2 the flash transform of x2 is equal to x1 and 0 which is 0 and the second equation we'll have s with flash transform of x2 plus 2 to the floss transform we're going to move this or shift these two we're going to have 2 plus s the plosh transform of x1 plus s the plosh transform of x2 is equal to x2 of 0 which is 1 and now we can write this as a matrix equation so we'll have s negative 2 2 and s times a vector of x1 and x2 the plosh transforms is equal to 0 and 1 and now we can use kramer's rule to write that the loplas transform of x1 is equal to the determinant of b1 over the determinant of a, where the determinant of b1 is equal to replacing b with the column with the first column in a.

01:43 So this would be 0 -1, negative 2, s, which is just 2.

01:50 We're going to calculate the determinant of a, which is the determinant of s, negative 2, 2, and s, which will give us s squared plus 4.

02:02 So therefore this is 2 over s squared plus 4.

02:07 We're going to take the inverse of plosh transform to just get back to our x1 of t, which will be the inverse of plosh transform of 2 over s squared plus 2 squared.

02:19 And we can use the rule to have the little plosh transform of the sign of bt is equal to b over s squared plus b squared, which is exactly what we have up here.

02:31 With the b being equal to 2...

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