Initial value problem

**Rapha** · November 11th 2008, 03:18 AM

Hi.

Solve the Initial value problem:

$\dot{u}(t) = \begin{pmatrix} a & 1 \\ 0 & b \end{pmatrix}\cdot u(t) \ , \ t\in [0,T] \ , \ u(0) = (1,1)^t$

(theres nothing said about 'a' and 'b')

does anyone know how to solve it, because i dont

Thanks so much!

Rapha

**Opalg** · November 11th 2008, 08:16 AM

Originally Posted by Rapha

Hi.

Solve the Initial value problem:

$\dot{u}(t) = \begin{pmatrix} a & 1 \\ 0 & b \end{pmatrix}\cdot u(t) \ , \ t\in [0,T] \ , \ u(0) = (1,1)^t$

(there's nothing said about 'a' and 'b')

Let $u(t) = \begin{bmatrix}x(t)\\y(t)\end{bmatrix}$ . Then $\begin{bmatrix}\dot{x}(t)\\\dot{y}(t)\end{bmatrix} = \begin{bmatrix}a & 1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatri x}$ . This says that $\dot{x} = ax+y,\ \dot{y} = by$ , with initial conditions x(0)=y(0)=1. The equation for y should be easy to solve. Then substitute the formula for y into the equation for x. Again, it will be a reasonably easy ordinary differential equation (but there will be two different forms of the solution, depending on whether a=b or not).

**Rapha** · November 12th 2008, 07:53 AM

Hey Opalg,

I really appreciate your answer, thank you

Originally Posted by Opalg

Let $u(t) = \begin{bmatrix}x(t)\\y(t)\end{bmatrix}$ . Then $\begin{bmatrix}\dot{x}(t)\\\dot{y}(t)\end{bmatrix} = \begin{bmatrix}a & 1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatri x}$ . This says that $\dot{x} = ax+y,\ \dot{y} = by$ , with initial conditions x(0)=y(0)=1. The equation for y should be easy to solve.

Indeed, $y(t) = e^{bt} \Rightarrow \dot{y} = y'(t) = b*e^{bt}$

Originally Posted by Opalg

Then substitute the formula for y into the equation for x. Again, it will be a reasonably easy ordinary differential equation (but there will be two different forms of the solution, depending on whether a=b or not).

$\dot{x} = ax(t)+e^{bt}$

I dont get it

**Opalg** · November 12th 2008, 10:42 AM

Originally Posted by Rapha

$\dot{x} = ax(t)+e^{bt}$

I don't get it

If a ≠ b then you should be able to find a solution of the form $x(t) = Ae^{at} + Be^{bt}$ , for suitable constants A and B.

If a = b then the solution will be of the form $x(t) = (At+B)e^{at}$ .

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