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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 $u(t)+e(t)=y(t)$?

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  • $\begingroup$ Two points. First, at least part of your problem about showing $e(t)$ satisfies the IVP is trivial. You should at least spot which of the two parts is easier. Second, the last part of the Question assumes $f(t,y) = \lambda y$. This lack of dependence of $f$ on "time" $t$ is often referred to as an autonomous system. Special deductions can be made from this property. $\endgroup$ – hardmath Apr 15 '16 at 2:26
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For the first question, notice that \begin{align*} e' &= (y-u)' \\ &= y' - u' \\ &= f(t,y) - u' \tag{1}. \end{align*} In equation (1) we can use $y=u+e$. This gives $$ e' = f(t,u+e)-u'. $$

For the second question, if $y'=\lambda y$, then using separation of variables gives $y(t) = y_0 e^{\lambda t}$. Since $f(t,y) =\lambda y$, the differential equation for $e$ takes the form \begin{align*} e' &= f(t,u+e) - u' \\ &= \lambda (u+e)-u'. \end{align*} Put $e$ on the left hand side, and $u$ on the right hand side. This gives $$ e' -\lambda e = - (u'-\lambda u). $$ Using the integrating factor $\mu(t) = \exp{(-\lambda t)}$, we obtain $$ (\mu(t) e )' = - (\mu(t) u)' \implies \int (\mu(t) e )'\,dt =- \int (\mu(t) u)'\,dt. $$ Hence, $\mu(t) e = -\mu(t) u +C$, where $C$ is a constant of integration. Using the initial condition $e(0)=0$, one finds that $C=y_0$. Solving for $e$ then gives $$ e(t) = -u(t) + \frac{y_0}{\mu(t)}. $$ Notice that $1/\mu(t) = \exp{( \lambda t)}$, so $$ e(t) = -u(t) + \frac{y_0}{\mu(t)} = -u(t) +y_0\exp{( \lambda t)} = -u(t)+y(t). $$ Thus, $u(t)+e(t) =y(t)$.

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