Which of the following is the solution to the differential equation dy/dx = e^{(y + x)} with initial condition y(0) = -ln4
y = -x - ln4
y = x - ln4
y = -ln(-e^x + 5)
y = -ln(e^x + 3)
y = ln(e^x + 3)

Solution:

Given: The differential equation dy/dx = e^{(y + x)} with initial condition y(0) = -ln4

This is in the form x^{m + n} = x^m . xⁿ

So, e^{x + y} = e^x . e^y

dy/dx = e^x . e^y

By variable separable method, we put x terms one side and y terms on other side

1/e^y .dy = e^x .dx

e^-y . dy = e^x .dx

e^-y /-1 = e^x + c

-e^-y = e^x +c

e^x + e^-y = c

When x = 0, y = -ln4

e⁰ + e^-(-ln4) = c

e⁰ + e^ln4 = c

1 + 4 = c

c = 5

Now, e^x + e^-y = 5

e^-y =  5 - e^x

e^-y = - e^x+5

-y = ln(-e^x + 5)

y = -ln(-e^x + 5)

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Which of the following is the solution to the differential equation dy/dx = e^{(y + x)} with initial condition y(0) = -ln4

Summary:

The following is the solution to the differential equation dy/dx = e^{(y + x)} with initial condition y(0) = -ln4 , y = -ln(-e^x + 5)