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$\begingroup$ This is not a duplicate of Show that {xy∣|x|=|y|,x≠y} is context-free. The marker # separating x and y makes the problem simpler. Furthermore, the referenced problem assumes |x|=|y|, which is not the case here. It is true that the answer to the older question could be used...

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How do you solve y'+3y=0 given y(0)=4? | Socratic

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y = 4e^(-3x) We have y'+3y=0 , or: dy/dx + 3y = 0 This is a first order linear ordinary differential equation and so we can derive an Integrating Factor using: IF = e^(int 3dx)

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