Solve 0=u''+u*e^x | Poincare Return Map

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How do I write u''+u*e^x = 0 as a planar system?
 
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Okay its:

Let z=u', then z'=u''=-exp(x)*u.

So the planar system is {u'=z,z'=-exp(x)*u}, or in matrix form, U'=A*U, where U is the column vector (u,z) and A is the 2x2 matrix (0,1,-exp(x),0).

What can be said about the Poincare return map on the transversal u>0?
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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