Question

In Exercises \(1-10\) , use separation of variables to solve the initial value problem. Indicate the domain over which the solution is valid. \(\frac{d y}{d x}=(\cos x) e^{y+\sin x} \quad\) and \(y=0\) when \(x=0\)

Short answer

The solution to the initial value differential equation is \( y = -\sin x - \ln(-1 - \sin x) \) and it's valid in the domain \( x \in (\frac{3\pi}{2}, 2\pi) \) union \( (2n\pi, (2n+1)\pi) \) for any integer n.

Step-by-step solution

Arrange the Equation

Firstly, the given differential equation is rearranged so as to have variables 'y' and 'x' separated on both sides of the equation. Like this: \( dy = (\cos x) e^{y+\sin x} dx \)

Simplify the Equation

Next, divide both sides by \( e^{y+\sin x} \), which separates the variables completely, yielding \( e^{-\sin x - y} dy = \cos x dx \)

Integration

Now we integrate both sides of the equation: \( ∫e^{-\sin x - y} dy = ∫\cos x dx \) which yields \( -e^{-\sin x - y} = \sin x + C \) where C is the constant of integration.

Solve for y

Solve the above equation for y to modify it into the form y = f(x) by bringing -y to one side of the equation and other terms to the other side. We obtain \( y = -\sin x - \ln(-C - \sin x) \)

Apply the Initial Condition

To evaluate the constant C, substitute y = 0 and x = 0 into the equation obtained in Step 4. This leads to: \( 0 = 0 - \ln(-C) \), from which we get \( C = -1 \)

Final Solution

Substitute C = -1 back into the equation obtained in step 4. The final solution to the differential equation is \( y = -\sin x - \ln(-1 - \sin x) \).

Determine the Validity Domain

Lastly, determine the domain where this solution is valid. We already know that logarithm function is defined for positive arguments, so the domain of the solution has to satisfy: -1 - \sin x > 0. Solving this inequality gives the domain as \( x \in (\frac{3\pi}{2}, 2\pi) \) union \( (2n\pi, (2n+1)\pi) \) where n is any integer.