Heaviside Function: What is H(-x)? Fourier Transform

squenshl
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What is H(-x)? Is it 1 if x < 0, 0 if x \geq 0. If so what is the Fourier transform of H(-x)exp(x)?
 
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By definition, H(x) is 0 if x < 0, 1 if x > 0, so setting x -> -x gives H(-x) is 0 if x > 0, 1 if x < 1. By inspection, you should be able to relate H(-x) to H(x), by H(-x) = 1 - H(x), which will help you with your Fourier transform.
 
Sweet.
My Fourier transform is f(w) = 1/(1-iw)
 
So is the Fourier transform of H(x)exp(-2x) + H(-x)exp(x)
f(w) = 1/(2+iw) + 1/(1-iw)?
 
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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