Let $ε, δ > 0$ be given. Let $φ$, $ψ$ be solutions of the two initial value problems $$y'' + y = xy ,y(0) = 1 , y'(0) = 2$$ , and $$y'' + y = xy + ε, y(0) = 1 + δ , y'(0) = 2 + δ$$ respectively on an interval $I := [0, a]$ for some $a > 0$. Then show that $$|φ(x) − ψ(x)| ≤ δ(1 + a)e^{a(a+1)x}+\frac{ε}{a + 1} [e^{a(a+1)x − 1}],∀ x ∈ I$$ .
Now, When I convert the second order odes to a system of first order ode, and apply continuous dependence estimate, I'm getting $$|φ(x) − ψ(x)| ≤ δe^{(a+2)x}+\frac{ε}{a + 2} [e^{(a+2)x − 1}],∀ x ∈ I$$
Is there some other theorem which can be used to prove this inequality? Is there a way to reach the required inequality from what I am getting?
So the idea for the perturbation estimate is that if $u'=f(x,u)$ and $v'=f(x,v)+εn$ where $f,u,v,n$ are vector-valued and the Lipschitz constant of $f$ is $L$, then \begin{align} \|u(x)-v(x)\| &\le\|u(0)-v(0)\|+\int_0^x\|f(s,u(s))-f(s,v(s))-εn\|ds \\ &\le\|u(0)-v(0)\|+\int_0^x(L\|u(s)-v(s)\|+ε\|n\|)\,ds \\[1em]\hline \implies (L\|u(x)-v(x)\|+ε\|n\|)&\le (L\|u(0)-v(0)\|+ε\|n\|)e^{L|x|} \end{align} Now take $u=(φ,φ')$, $v=(ψ,ψ')$ as state vectors and the maximum norm as the vector norm on the state space. This gives $f(x,u)=(φ', (x-1)φ)$ as the ODE function, which results in, as far as I see, $L=\max(1,a-1)$ as Lipschitz constant on $[0,a]$. $\|n\|=\|(0,1)\|=1$ and $\|u(0)-v(0)\|=\|(-δ,-δ)\|=δ$ then gives the remainder of the used parameters. $$ |\phi(x)-\psi(x)|\le δe^{L|x|}+\frac{ε}{L}(e^{L|x|}-1) $$ This appears to be a tighter bound than both proposed versions.