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?
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.