Calculus

Find the particular solution (solved for y) for the differential equation dy/dx=2x/e^(2y) satisfying y(0)=1.

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  1. e^-2y dy = 2 x dx

    e-2y /-2 = x^2 + c

    e^-2y = -2 x^2 + c

    -2y = ln (-2 x^2+c)

    when x = 0, y = 1
    -2 = ln(c)
    e^-2 = c

    -2 y = ln (-2x^2+e^-2)

    y = -(1/2) ln(-2x^2+1/e^2)

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    posted by Damon
  2. Can you explain why it's simplified to e^-2ydy=2xdx. I don't understand why it's e^negative 2y

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