Question

which of the following is the solution to the differentiable equation dy/dx=e^y+x with the initial condtion y(0)=-ln(4)

Answer 1

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

Explanation:

I assume you mean:

dy/dx = e^(y + x)

Use exponent properties:

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

Separate the variables:

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

Integrate:

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

Solve for y:

e^(-y) = -e^x − C

-y = ln(-e^x − C)

y = -ln(-e^x − C)

Use initial condition to solve for C:

-ln 4 = -ln(-e^0 − C)

4 = -1 − C

C = -5

Therefore, the equation is:

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