Mathematical modeling: exact solution

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Let $E_0 > 0$. For a parameter $E \in [0, E_0] $, I consider the family of operators:

$ F( \cdot , E) : C_b^2 ([0, \infty)) \to C_b^0 ([0, \infty)) \times \mathbb R^2$

$ F(y, E) = (y^n + (1+E)y, y(0), y'(0)-1)$.

$C_b^n([0, \infty))$ is the space of the n-times differentiable functions, where the functions and the derivations are limited to the order n. The norms on $C_b^2([0, \infty))$ and $C_b^0([0, \infty)) \times \mathbb R^2$ are given by:

$ ||y||_{C_b^2([0, \infty))} = sup (|y(t)| + |y'(t)| + |y''(t)| )$

$ ||(f, a, b)||_{C_b^0([0, \infty)) \times \mathbb R^2} = sup (|f(t)|) + |a| + |b| $

(where $t \in (0, \infty)$)

Now my question is: what is the $exact$ solution for $F(y, E) = (0, 0, 0)$?

Thanks for any hint / help!


Edit: What I have to solve, at the end, is:

$y'' + \alpha^2 y = 0$ , where $\alpha = \sqrt{1+ E}$

with $y(0) = 0$ and $y'(0)=0$.

Can anybody help?


Since E > 0, we have: $e^{\alpha x} (c_1 cos(\beta x) + c_2 sin(\beta x))$

We know: y(0) = 0 --> $c_1 = 0$

y(0)' = 1: $e^{\alpha x}(\beta c_2 cos(\beta x) - \beta c_1 sin (\beta x)) + \alpha e^{\alpha x} (c_1 cos(\beta x) + c_2 sin(\beta x))$ --> $\alpha c_1 + \beta c_2$ --> $\beta c_2 = 1$.

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Is simply the solution of the linear ODE $$y'' + (1 + E)y = 0$$ with the the initial conditions $$y(0) = 0,\qquad y'(0) = 1.$$ Can you continue?