# Differential equations and linear algebra way

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?

• What is $x$ in your post?
– user9464
Commented Feb 10, 2021 at 21:27
• If $u$ is a scalar, then $x$ is nothing but a number. Can you give more information? Where do you see this equation? What book are you reading?
– user9464
Commented Feb 10, 2021 at 21:30
• I am reading 6.3 Systems of Differential Equations by gilbert strang. the book said, Ax=lambda*x, so x should be eigenvector. Commented Feb 10, 2021 at 21:34
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.
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 $$\frac{du}{dt}=Au.$$