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

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

Step-by-step 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)