Solve the differential equation y dx+(2xy-e^(-2y))dy=0.

I know that the integrating factor is e^2y/y.

e^2y dx+2x(e^2y)-1/y dy=0

Now what?

The answer is xe^2y-ln abs(y)=c, y=0. Can you explain how to get y=0?

Solve the differential equation y dx+(2xy-e^(-2y))dy=0.

I know that the integrating factor is e^2y/y.

e^2y dx+2x(e^2y)-1/y dy=0

Now what?

The answer is xe^2y-ln abs(y)=c, y=0. Can you explain how to get y=0?

Tutors, please sign in to answer this question.

Let x' = dx/dy. The equation can be changed to,

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

So, this is a typical first order linear differential equation, which has a general solution of

x = (1/e^2y)[∫e^2y (1/y)e^(-2y) dy + c] = (1/e^2y)[ln|y| + c]

Multiply both sides by e^2y and simiplify,

xe^2y - ln|y| = c, where y ≠ 0 because if y = 0, x' + 2x = (1/y)e^(-2y) is undefined.

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Attn: Here y is a variable, x is the function of y.

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