Differential equations and linear algebra way

Question

I am a beginner at differential equations. Never learned this before but do know calculus.

I am reading Section 6.3 (Systems of Differential Equations) in Introduction to Linear Algebra by Gilbert Strang.

At the beginning of the section:

So, e^ut is equal e^lambdatx. I don't understand this part.

So, here are what I understand:

$$ \frac{\mathrm{d}u }{\mathrm{d} t} = u(t) = {e^{ut}} $$

And in the linear algebra:

It turns out,

$$ u(t) = e^{\lambda t}x $$

Can anyone help to explain or let me know the key word to find the relation?

Answer 1

One way is separation of variables. One can prove that the solution with $u(0) = x$ (in your notation) of $\frac{\mathrm{d}}{\mathrm{d}t}u(t) = u(t)$ fulfills $$ \int^{u(t)}_x \frac{1}{g(s)}~\mathrm{d}s= \int^t_0 h(s)~\mathrm{d}s, $$ where $g(s) = s$ and $h(s) = 1$. This comes from the ODE being in the form $$ \frac{\mathrm{d}}{\mathrm{d}t} u(t) = g(u)h(t). $$ I'm sure you can find the solution yourself.

Answer 2

If $Ax=\lambda x$, and $u(t)=e^{\lambda t}x$, then direct substitution tells you that $$ \frac{du}{dt}=\frac{d}{dt}e^{\lambda t}x=\lambda e^{\lambda t}x=e^{\lambda t}(\lambda x)=e^{\lambda t}(Ax)=A(e^{\lambda t}x)=Au $$

Therefore, $u(t)=e^{\lambda t}x$ solves that differential equation $$ \frac{du}{dt}=Au. $$