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How do you find the general solution to #dy/dx+e^(x+y)=0#?

Answer 1

Paisley Johnson

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

#dy/dx+e^(x+y)=0#

this is separable

#dy/dx= - e^(x+y)#

#dy/dx= - e^x e^y#

#e^(-y) dy/dx= - e^x#

#int\ e^(-y) dy/dx \ dx=int - e^x \ dx#

#int\ e^(-y) \ dy=- int e^x \ dx#

#-e^(-y) =- e^x + C#

#e^(-y) = e^x + C#

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

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

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Answer 2

Emma Aldridge

To find the general solution to the differential equation ( \frac{dy}{dx} + e^{x+y} = 0 ), first separate the variables by moving terms involving ( y ) to the left side and terms involving ( x ) to the right side. Then integrate both sides with respect to their respective variables. The solution will involve an arbitrary constant ( C ). The steps are as follows:

Rearrange the equation to separate variables:

[ \frac{dy}{dx} = -e^{x+y} ]

Integrate both sides:

[ \int dy = \int -e^{x+y} dx ]

Integrate with respect to ( y ) on the left side and with respect to ( x ) on the right side:

[ y = -\int e^{x+y} dx + C ]

Integrate ( -e^{x+y} ) with respect to ( x ). This might require using techniques such as substitution or integration by parts.
After integrating, solve for ( y ) to obtain the general solution.

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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.