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What is a solution to the differential equation #y'=-2xe^y#?

Answer 1

Cameron Alexander

#y = ln(1/(x^2 + C))#

#y'=-2xe^y#

this is separable

# e^-y \ y' =-2x#

#int e^-y \ y' \ dx =-2 int x \ dx#

#int \ e^-y \ dy =-2 int x \ dx#

# -e^-y \ =-2 ( x^2/2 + C)#

# e^-y \ = x^2 + C#

# e^y \ = 1/(x^2 + C)#

# ln (e^y) \ = ln(1/(x^2 + C))#

#y = ln(1/(x^2 + C))#

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Answer 2

Cameron Alexander

The solution to the differential equation ( y' = -2xe^y ) can be obtained using separation of variables. After separating variables and integrating both sides, the solution is:

[ y = -\ln(|1 - x^2|) + C ]

where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.