Eigenvalue and system solution

Question

Prove that if $\lambda$ is an eigenvalue associated to the constant vector $v \in \mathbb R^n$ of the matrix A(t) then $ x(t) = \exp(\int_0^t(\lambda(y)dy))v$ is a solution of the system $dx/dt = A(t)x$

I started by showing that if $\lambda$ is an eigenvalue associated to the matrix A then we have that $ x(t) = \exp(\int_0^t(\lambda(y)dy))v$ has a derivative equal to what is required.

Answer 1

$$ x(t) = \exp\left(\int_0^t\lambda(y)dy\right)v$$ Taking the derivative with respect to $t$ you get $$\frac{dx}{dt}=\lambda(t)\exp\left(\int_0^t\lambda(y)dy\right)v$$ We can use the fact that $\lambda(t)$ and the exponential are just some numbers (functions of $t$), to write $$\begin{align}\frac{dx}{dt}&=\exp\left(\int_0^t\lambda(y)dy\right)\lambda(t)v\\&=\exp\left(\int_0^t\lambda(y)dy\right)A(t)v\\&=A(t)\exp\left(\int_0^t\lambda(y)dy\right)v\\&=A(t)x(t) \end{align}$$

Answer 2

You have that $dx/dt = exp(\int_0^t(\lambda(y)dy)[\lambda(t)]v=\lambda(t)*x(t)$.

Using the fact that $\lambda$ is an eigenvalue of the matrix A then you have that $dx/dt = A(t)x(t)$