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

Inverse Laplace Transform

After finding the algebraic solutions in the Laplace domain, the inverse Laplace transform is used to convert these solutions back into the time domain. Techniques used in computing the inverse transform include partial fraction decomposition and the application of known inverse transform formulas, ultimately yielding the time-domain functions that satisfy the original differential equations.

Algebraic Manipulation in the Laplace Domain

Once the Laplace transform is applied, the resulting algebraic equations are manipulated to solve for the transforms of the unknown functions. This approach often involves combining like terms, factoring, and employing methods such as substitution to decouple the equations in order to isolate each function's Laplace transform.

Laplace Transform

The Laplace transform is a mathematical technique that converts differential equations in the time domain into algebraic equations in the complex frequency domain. It is especially powerful for linear ordinary differential equations with constant coefficients because it simplifies differentiation into multiplication by a complex variable, making the equations easier to manipulate and solve.

Initial Value Problems

Initial value problems involve differential equations along with specified values of the unknown function and its derivatives at a starting point. These initial conditions are incorporated into the Laplace transform process, where the transform of derivatives includes terms based on the initial values, enabling the unique determination of the solution.

Systems of Differential Equations

Systems of differential equations consist of multiple interrelated equations that describe the behavior of several unknown functions simultaneously. In such systems, methods like the Laplace transform are applied to each equation, thereby converting a system of differential problems into a system of algebraic equations that can be solved using linear algebra techniques.

Transcript

00:01 The first thing we're going to do here is take our second initial condition and plug it into the first equation that we have.

00:08 So x1 prime at 0 is equal to 0 is equal to x1 of 0 plus 2 x2 of 0.

00:19 And we know the x1 of 0 is 1.

00:23 So therefore x2 of 0 is negative 1 half.

00:30 Now we're going to use this later on, but we're going to use this later on, but we're going to and take the laplace transform of both of our equations to get that this is s times l 'plas transform of x1 minus x1 at 0 minus the laplace transform of x1 minus 2 times the plus transform of x2 is equal to 0 as well as x or s times the flosh transform of x2 minus x2 at 0 minus 2 the flash transform of x1 minus the plosh transform of x2 is equal to 0 and we're going to group terms together by factors of x1 and x2 so that this becomes s minus 2 times x1 should be x minus or s minus 1 times x1 minus 2 x2 is equal to x1 and 0 which is 1 and 0 which is 1 and we're going to going to have a negative 2 x1 plus s minus 1 times x2 is equal to x2 of 0 which is negative 1 1 1⁄ and now we have a system of equations that we can write as a times x is equal to b where x and b are vectors and a is a matrix so our a is going to be s minus 1, negative 2, negative 2, and s minus 1.

02:24 Our x is the plus transform of x1 and the plus transform of x2.

02:30 And our b is just 1, negative 1 1⁄2, the right side here.

02:38 And now we can use kramer's rule to figure out what x1 and x2 are.

02:42 So we know that x1 will be the determinant of e1.

02:50 Over the determinant of a, where the determinant of a will be the determinant of s minus 1, negative 2, negative 2, and s minus 1, which will be s minus 1 squared, plus 4, which we can expand to be s squared minus 2s plus 1, minus 4.

03:20 It should be a minus 4, not plus, since negative 2 times negative 2 is positive.

03:24 And we have to subtract the off diagonal multiplication, which is s squared minus 2 s minus 3, which is s minus 3, and s plus 1.

03:37 So there we have the determinant of a.

03:39 Now we're going to calculate the determinant of b1, which is equal to replacing the first column in our a with b.

03:52 So this will be 1 negative 1 half, negative 2 s minus 1 and we calculate this determinant it will be s minus 1 minus 1 minus negative 2 times 1 half is just 1 so it just be s minus 2 so therefore x1 is s minus 2 or laplace transform of x1 is s minus 2 over s minus 3 times s plus 1 and we're going to write this as two separate fractions 1 for s minus 3 and 1 for s plus 1 and to do that we're going to multiply through by the denominator to get that s minus 2 has to be equal to a times s plus 1 plus b times s minus 3 and foiling out the terms we'll get that a plus b all times s plus a minus 3 b has to be able to s minus 2 so therefore a plus b has to be equal to 1 or a is equal to 1 minus b and in this second equation we have a minus 3b you can make the substitution that a is 1 minus b so 1 minus b minus is a negative 4b is equal to negative 2 so therefore b is 3 quarters if b is 3 quarters then a is 1 minus 3 quarters, which is negative 1 quarter.

05:49 No, positive 1 quarter...

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