Suppose that $y(t)$ is the exact solution of the ivp $$y'(t)=f(t,y(t)), y(0)=y_0$$ and $u(t)$ is any approximation to $y(t)$ with $u(0)=y(0)$. Define the error $e(t)=y(t)-u(t)$.
How can I show that $e(t)$ satisfies the ivp $$e'(t)=f(t,u(t)+e(t))-u'(t), e(0)=0$$ And if we say that $f(t,y)=\lambda y$ for some constant $\lambda$, how can we solve the ivp from my previous question to show that $u(t)+e(t)=y(t)$?