4. Solve the initial value problem The matrix A =

4. Solve the initial value problem The matrix $A = \begin{bmatrix} 1 & 1 \\ 3 & 1 \end{bmatrix}$ has eigenvalues $\lambda = 1 + \sqrt{3}$, $\mu = 1 - \sqrt{3}$, and the general solution to $Dx = Ax$ is given by $\mathbf{x}(t) = c_1 e^{\lambda t} \begin{bmatrix} 1 \\ \sqrt{3} \end{bmatrix} + c_2 e^{\mu t} \begin{bmatrix} -1 \\ \sqrt{3} \end{bmatrix}$. a. Find the general solution to the inhomogeneous equation $Dx = Ax + \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. b. Solve the initial value problem $Dx = Ax + \begin{bmatrix} 1 \\ 1 \end{bmatrix}$; $\mathbf{x}(0) = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

Transcript

00:01 Hello students, in this question we have to solve the initial value problem that is dx by dt equal to ax with the condition x of 0 equal to x0.

00:15 So that here the given matrix is a equal to 1 by 2 minus 7 by 2 minus 7 by 2 and 1 by 2 with here it is given that x0 is 3 4.

00:30 So that first we have to find the eigenvalues, eigenvectors and the coefficients c1 and c2.

00:35 First to find the eigenvalues we get a determinant of a minus lambda i equal to 0 that is determinant of 1 by 2 minus lambda minus 7 by 2 minus 7 by 2 1 by 2 minus lambda equal to 0.

00:53 Therefore it will be 1 by 2 minus lambda into 1 by 2 minus lambda minus minus 7 by 2 into minus 7 by 2 equal to 0.

01:05 By solving this we will get 1 by 2 minus lambda the whole square equal to 49 by 4.

01:12 Here we can take square root on both sides that we will get 1 by 2 minus lambda equal to plus or minus 7 by 2.

01:21 By solving this we get lambda equal to minus 3 comma 4.

01:29 Therefore the eigenvalues are minus 3 and 4.

01:42 For eigenvectors we know that a minus lambda into i into x equals to 0.

01:54 Here for lambda equal to minus 3 we will have a plus 3 i into x equal to 0 such that we will have here a plus 3 i is 7 by 2 minus 7 by 2 minus 7 by 2 and 7 by 2 into here x can be written as x1 x2 equal to 0.

02:20 Such that we will get by matrix multiplication that is 7 by 2 x1 minus 7 by 2 x2 equal to 0 and minus 7 by 2 x1 plus 7 by 2 x2 equal to 0.

02:37 Such that we can solve this it will be 7 by 2 x1 equal to 7 by 2 x2 such that x1 equal to x2.

02:50 Hence we can say that the eigenvector is here let us check eigenvector v1 which is equal to 1 1.

03:01 Next we have to find the eigenvector at lambda equal to 4...

Question

Please give Ace some feedback