Question

Solve the given initial-value problem. x' = 10 −1 5 8 x, x(0) = −6 8

Answer 1

Due to a typographical error, a general solution method for a first-order linear differential equation and handling of an initial condition is suggested; specific values are needed for an exact solution.

Step-by-step explanation:

To solve the initial-value problem x' = 10 − 1 5 8 x, where x(0) = −6 8, it seems there may have been a misunderstanding or typographical error in the presentation of the problem. Therefore, I will provide a general approach to solving a first-order linear differential equation that resembles the one given, and how to handle an initial condition.

Typically, for a first-order linear differential equation in the form dx/dt = ax + b, where 'a' and 'b' are constants, the solution can be found using the method of integrating factors or by separation of variables if the equation can be rewritten appropriately.

Once the general solution is derived, the initial condition x(0) = C (where C is a given constant) is used to find the particular solution, by substititing the initial value into the general solution and solving for the constant of integration.

Note: The specific values must be clear to provide an exact solution;

However, without clear coefficients, it is not possible to offer a precise solution to the equation.

Answer 2

It looks like this is a system of linear ODEs given in matrix form,

`x' = \begin{bmatrix}10&-1\\5&8\end{bmatrix} x`

with initial condition x(0) = (-6, 8)ᵀ.

Compute the eigenvalues and -vectors of the coefficient matrix:

`\det\begin{bmatrix}10-\lambda&-1\\5&8-\lambda\end{bmatrix} = (10-\lambda)(8-\lambda) + 5 = 0 \implies \lambda^2-18\lambda+85=0 \implies \lambda = 9\pm2i`

Let v be the eigenvector corresponding to λ = 9 + 2i. Then

`\begin{bmatrix}10-\lambda&-1\\5&8-\lambda\end{bmatrix}v = 0 \implies \begin{bmatrix}1-2i&-1\\5&-1-2i\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}`

or equivalently,

`\begin{cases}(1-2i)v_1-v_2=0 \\ 5v_1-(1+2i)v_2=0\end{cases} \implies 5v_1 - (1+2i)v_2 = 0`

Let
`v_2=1-2i`; then
`5v_1-5=0\implies v_1=1`, so that

`\begin{bmatrix}10&-1\\5&8\end{bmatrix}\begin{bmatrix}1\\1-2i\end{bmatrix} = (9+2i)\begin{bmatrix}1\\1-2i\end{bmatrix}`

and we get the other eigenvalue/-vector pair by taking the complex conjugate,

`\begin{bmatrix}10&-1\\5&8\end{bmatrix}\begin{bmatrix}1\\1+2i\end{bmatrix} = (9-2i)\begin{bmatrix}1\\1+2i\end{bmatrix}`

Then the characteristic solution to the system is

`x = C_1 e^((9+2i)t) \begin{bmatrix}1\\1-2i\end{bmatrix} + C_2 e^((9-2i)t) \begin{bmatrix}1\\1+2i\end{bmatrix}`

From the given condition, we have

`\displaystyle \begin{bmatrix}-6\\8\end{bmatrix} = C_1 \begin{bmatrix}1\\1-2i\end{bmatrix} + C_2 \begin{bmatrix}1\\1+2i\end{bmatrix} \implies C_1 = -3-\frac i2, C_2=-3+\frac i2`

and so the particular solution to the IVP is

`\displaystyle \boxed{x = -\left(3+\frac i2\right) e^((9+2i)t) \begin{bmatrix}1\\1-2i\end{bmatrix} - \left(3-\frac i2\right) e^((9-2i)t) \begin{bmatrix}1\\1+2i\end{bmatrix}}`

which you could go on to rewrite using Euler's formula,

`e^((a+bi)t) = e^(at) (\cos(bt) + i \sin(bt))`