Find the particular solution of the linear system that satisfies the stated initial conditions.

Find the particular solution of the linear system that...

Key Concepts

Initial Value Problem

An initial value problem specifies the state of the system at a particular time, usually t = 0, and the goal is to find the particular solution that satisfies these initial conditions. Incorporating the given initial conditions often involves solving for unknown constants in the general solution derived from methods like the matrix exponential.

Eigenvalues and Eigenvectors

A central approach to solving linear systems involves finding the eigenvalues and eigenvectors of the coefficient matrix. These quantities allow the system to be diagonalized, which simplifies the process of finding the general solution by decoupling the system into independent scalar differential equations.

Matrix Exponential Method

Once the system is written in matrix form and the eigen decomposition is obtained, the solution can be expressed in terms of the matrix exponential. This method directly incorporates the effects of the coefficient matrix over time and is especially useful in dealing with initial value problems.

Coefficient Matrix

The system can be expressed in matrix form where the derivatives are given by the product of a coefficient matrix and a vector of variables. This formulation enables the use of linear algebra techniques, such as finding eigenvalues and eigenvectors, to analyze and solve the system.

Linear Systems of Differential Equations

These systems consist of multiple differential equations where each derivative is expressed as a linear combination of the variables. The methods for solving such systems often involve techniques such as decoupling the equations or using matrix methods to reduce the system to simpler, solvable parts.

Transcript

00:01 Dear students, here we have a question and we are asked to find the particular solution of the linear system that satisfies the initial condition.

00:08 So our linear systems are the x upon d t is equal to 7x minus y and d y upon d t is equal to 4x plus 3y.

00:16 So our initial condition for x of 0 is 1 and y of 0 is equal to 3.

00:21 So as we know that, d x upon d t is a differential operator and we can write it in the form of d into x.

00:31 So, we'll write dx and shift all the terms on the left hand side, so we'll get minus 7x plus y equals to 0.

00:41 This will be our first equation and dy and shift all the terms on the left inside.

00:49 So we'll get minus 4x minus 3y equals to 0.

00:53 This will be our second equation.

00:56 So if you take x common, we'll get d minus 7 into x plus y equals to 0.

01:03 And if we take y common we'll get minus 4 x plus d minus 3 of y equals to 0 so we can solve these equations by kramer's rule as in kramer's rule you first need to find del del del is operator we can find it by taking the determinant of our two equations so the coefficients of x is d minus 7 and y is 1 and for the second equation the coefficient of x is minus 4 and the coefficient of y is d minus 3 so our determinant will be we will multiply the coefficients d minus 7 d minus 3 and by multiplying these the other coefficients the other diagonals will get plus 4 so let's simplify it we'll get t square minus 70 minus 3d and by multiplying minus 7 with minus 3 will get plus of 21 plus 4.

02:12 So we'll get d squared minus 10 d and plus 25.

02:20 This is our del.

02:21 Now we need to find del x.

02:24 So our del x will be, we have got zeros on our right inside...

Please give Ace some feedback