# Initial value problem

• Nov 11th 2008, 04:18 AM
Rapha
Initial value problem
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
• Nov 11th 2008, 09:16 AM
Opalg
Quote:

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).
• Nov 12th 2008, 08:53 AM
Rapha
Hey Opalg,

Quote:

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

Quote:

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
• Nov 12th 2008, 11:42 AM
Opalg
Quote:

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 $x(t) = (At+B)e^{at}$.