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Find the Solution to the Initial Value P y 2 x y = e x 2 , y ( 0 ) = 1

Find the Solution to the Initial Value P y2xy=ex2,y(0)=1y2xy=ex2,y(0)=1 2 answers

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The given differential equation is y 2xy ex2 y0 1 This is a firstorder linear differentia... blur-text-image

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