What is the general equation of #dy/dx =e^(x+y)#?

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Answer 1 · 2018-04-05T08:42:47

#y = ln(1/(C_0-e^x))#

This is a separable differential equation so

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

#-e^-y= e^x - C_0# or

#y = ln(1/(C_0-e^x))#

Answer 2 · 2018-04-05T11:31:42

# y = -ln|C-e^x| #

We have:

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

Which we can write as

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

We can collect terms for similar variables:

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

Which is a separable First Order Ordinary non-linear Differential Equation, so we can "separate the variables" to get:

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

Both integrals are those of standard functions, so we can use that knowledge to directly integrate:

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

And we can readily rearrange for #y#:

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

# :. -y = ln|C-e^x| #

Leading to the General Solution:

# y = -ln|C-e^x| #