Problem 4-(20 P) Solve the given system of differential...

Problem 4-(20 P) Solve the given system of differential equations \begin{cases} \frac{dx}{dt} = 3x - 2y\\ \frac{dy}{dt} = -x + 3y - 2z\\ \frac{dz}{dt} = -y + 3z \end{cases} (\lambda_1 = 1, \lambda_2 = 3, \lambda_3 = 5)

Transcript

00:01 We are given dx by dt is equals to 3x minus 2y dy by dt is equals to minus x plus 3y minus 2z and we have dz by dt is equals to minus y plus 3z.

00:20 We have to solve the system of equations.

00:25 Solve the given system of equations.

00:31 So let us focus on our solutions.

00:32 So we let dx by dt.

00:35 So first of all, we have some assumptions.

00:39 So let x be the variable x1 and y be the variable x2 and z be the variable x3.

00:47 So we have dx by dt is equals to x1 dash, dy by dt is equals to x2 dash and dz by dt is equals to x3 dash.

01:02 So we can write the system of equation as x1 dash x2 dash and x3 dash.

01:09 This column vector is equals to we have the a matrix as 3 minus 2 0 minus 1 3 minus 2 0 minus 1 3 and here we have the column vector x1 x2 x3.

01:30 Let this be the matrix y.

01:33 Let this be a.

01:35 So let this be the matrix y.

01:37 So we have this as the matrix y dash.

01:40 So we have y dash is equals to a y.

01:44 Now we let y is equals to x multiplied by e raised to the power lambda t.

01:52 So this implies we have y dash is equals to lambda x multiplied by e raised to the power lambda t.

02:00 This implies we have lambda x is equals to a x.

02:04 Now we solve this system of equations.

02:06 So we have a minus lambda i times x is equals to 0.

02:12 So we have the matrix a minus lambda i determinant is equals to 0 and we get the eigenvalues from this equation.

02:21 So we have the determinant of 3 minus lambda minus 2 0 minus 1 3 minus lambda minus 2 times 0 minus 1 3 minus lambda.

02:34 So the determinant of this matrix is equals to 0.

02:39 Now let us find out the determinant.

02:40 So we have 3 minus lambda multiplied by 3 minus lambda square minus 2 plus 2 times 1 minus 0.

02:54 Here we have instead of 1 we have 2 times minus 1 multiplied by 3 minus lambda minus 0 minus 0 is equals to 0.

03:07 So we have 3 minus lambda cube minus 2 times 3 minus lambda minus 2 times 3 minus lambda is equals to 0.

03:17 So this implies we have 3 minus lambda multiplied by 3 minus lambda whole square minus 4 is equals to 0.

03:27 On solving this we get 3 minus lambda multiplied by.

03:30 So here we get 9 plus lambda square minus 6 lambda minus 4 is equals to 0.

03:37 So we have lambda is equals to 3 and lambda square minus 6 lambda plus 5 is equals to 0...

Question

Problem 4-(20 P) Solve the given system of differential equations \begin{cases} \frac{dx}{dt} = 3x - 2y\\ \frac{dy}{dt} = -x + 3y - 2z\\ \frac{dz}{dt} = -y + 3z \end{cases} (\lambda_1 = 1, \lambda_2 = 3, \lambda_3 = 5)

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