1. ## Three variables?

Write $\ddot x -2 \dot x -3x=10 \sin t$ in the three variables $x, \ y= \dot x$ and $\dot s =1$ so that the equations are autonomous and first order.
Clearly $\ddot x -2 \dot x -3x=10 \sin t$ is not a linear equation.

My lecture notes state that "nonlinear equations cannot be written in the form $\dot x = Ax$ (where $x \in \mathbb{R}^n$ and $A$ is an $n \times n$ matrix)".

So isn't this question literally asking me to do the impossible??

2. Originally Posted by Showcase_22
Clearly $\ddot x -2 \dot x -3x=10 \sin t$ is not a linear equation.

My lecture notes state that "nonlinear equations cannot be written in the form $\dot x = Ax$ (where $x \in \mathbb{R}^n$ and $A$ is an $n \times n$ matrix)".

So isn't this question literally asking me to do the impossible??
It's not asing for 3 linear equations just 3 equations.

Here
$\frac{ds}{dt} = 1$ so $\frac{dx}{dt} = \frac{dx}{ds}\cdot \frac{ds}{dt} = \frac{dx}{ds}$. Simiarly $\frac{dy}{dt} = \frac{dy}{ds}.$

Then $\frac{d^2x}{dt^2} = \frac{d}{dt} \left(\frac{dx}{dt}\right) = \frac{d}{ds} \left( \frac{dx}{ds}\right) \cdot \frac{ds}{dt} = \frac{d^2x}{ds^2} = \frac{dy}{ds}$ since $y = \frac{dx}{ds}$.

$
\frac{dt}{ds} = 1,
$

$
\frac{dx}{ds} = y,
$

$
\frac{dy}{ds} - 2 y - 3 x = 10 \sin t.
$

3. Ohhh, I see!

I have one more question though, how do I solve $\frac{dx}{ds}=y$ to get $x(t)$?

$y=\frac{dx}{dt}$ but it's completely wrong to put $x(t)=yt+C$ since $y$ is a function of t.

It's pretty easy to do the same for $\frac{dt}{ds}=1$ since this is equal to $\frac{ds}{dt}=1$ so $s(t)=t+C$.

I'm just a little bit stuck on what $x$ and $y$ actually are.

4. Originally Posted by Showcase_22
Ohhh, I see!

I have one more question though, how do I solve $\frac{dx}{ds}=y$ to get $x(t)$?

$y=\frac{dx}{dt}$ but it's completely wrong to put $x(t)=yt+C$ since $y$ is a function of t.

It's pretty easy to do the same for $\frac{dt}{ds}=1$ since this is equal to $\frac{ds}{dt}=1$ so $s(t)=t+C$.

I'm just a little bit stuck on what $x$ and $y$ actually are.
The middle step isn't right. $x$ and $y$ are unknown functions of $t$. You're treating y as constant. What you need to do is first solve the homogeneous system

$
\frac{dx}{ds} = y
$

$\frac{dy}{dx} = 3x + 2y$

5. $\frac{dx}{ds}=\frac{dx}{dt}=y$ and $\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{1} {y}.\frac{dy}{dt} =3x+2y$.

But the second equation is no longer a linear equation so it can't be put into matrix form.