1. ## Three variables?

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

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

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

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

My lecture notes state that "nonlinear equations cannot be written in the form $\displaystyle \dot x = Ax$ (where $\displaystyle x \in \mathbb{R}^n$ and $\displaystyle A$ is an $\displaystyle n \times n$ 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
$\displaystyle \frac{ds}{dt} = 1$ so $\displaystyle \frac{dx}{dt} = \frac{dx}{ds}\cdot \frac{ds}{dt} = \frac{dx}{ds}$. Simiarly $\displaystyle \frac{dy}{dt} = \frac{dy}{ds}.$

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

$\displaystyle \frac{dt}{ds} = 1,$
$\displaystyle \frac{dx}{ds} = y,$
$\displaystyle \frac{dy}{ds} - 2 y - 3 x = 10 \sin t.$

3. Ohhh, I see!

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

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

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

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

4. Originally Posted by Showcase_22
Ohhh, I see!

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

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

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

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

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

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

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