Boundary Value Problem + Green's Function

sassie
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Consider the BVP

y''+4y=f(x) (0\leqx\leq1)
y(0)=0 y'(1)=0

Find the Green's function (two-sided) for this problem.

Working: So firstly, I let y(x)=Asin2x+Bcos2x

Then using the boundary conditions,

Asin(2.0)+Bcos(2.0)=0 => B=0

y'(x)=2Acos(2x)-2Asin(2x)
y'(0)=2A=0 => A=0

But is this right? How can I derive a Green's function (two-sided) from this? Please help.
 
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Please ignore. I figured out what I did wrong.
 
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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