Aproximation to an ODE

Question

Consider the following Cauchy problem: $$\begin{cases} y'=y+\lambda x^2 \sin y \\y(0)=1 \end{cases}$$ Where $|\lambda|\le 1$. Show that if $\psi(x)$ is the solution, then $$|\psi(x) - e^x| \le | \lambda|(e^{|x|}-1)$$ For $|x|\le 1$ $$$$ I have tried computing Lipschitz constant (with respect to y) of $y+\lambda x^2 \sin(y)$ but the largest it gets is $1+\lambda x^2$. Also I have that $$|e^x - e^x-\lambda x^2 \sin(e^x)|\le \lambda x^2$$ So $e^x$ is a $\lambda x^2$-approximate solution to the problem. How can I manipulate these bounds to get the desired result?

Answer 1

Set $u(x)=\phi(x)-e^x$. You get $$ |(e^{-x}u(x))'|\le |λ|x^2e^{-x} $$ so that $$ \left|e^{-x}u(x)-u(0)\right|\le\int_0^x|(e^{-s}u(s))'|\,ds\le\int_0^x|λ|s^2e^{-s}\,ds $$ As $u(0)=0$ per initial conditions, after integration $$ |\phi(x)-e^x|=|u(x)|\le 2|λ|\,\Bigl|e^x-1-x-\tfrac12x^2\Bigr| $$ By manipulating the power series of the exponential you get $\frac{|λ|}3x^2(e^{|x|}-1)$ as a simplified greater bound.