(b) Find a particular solution of the differential equation $$(x+1) \frac{dy}{dx} = 2e^{-y} - 1, \text{ given that } y=0 \text{ when } x=0.$$ (differential equation, particular solution, initial condition, separable differential equation)

Step 1: Separate the variables

The given differential equation is a variable separable type. To solve it, we need to rearrange the terms so that all $y$ terms are on one side with $dy$, and all $x$ terms are on the other side with $dx$. We start by dividing both sides by $(2e^{-y}-1)$ and $(x+1)$ and multiplying both sides by $dx$.

Step 2: Integrate both sides

Now that the variables are separated, we integrate both sides of the equation. This step introduces an integration constant, which we will solve for later using the initial condition.

Step 3: Simplify the left integral

To integrate the left side, we multiply the numerator and denominator by $e^y$. This transforms the denominator into a simpler form for substitution, making the integral easier to solve.

Step 4: Substitute for the left integral

We use a substitution method for the left integral. Let $u = 2 - e^y$. Then, differentiate $u$ with respect to $y$ to find $du$. This substitution allows us to express the integral in terms of $u$.

Step 5: Evaluate the left integral

Substituting $u$ and $du$ into the left integral gives us $\int \frac{-du}{u}$. This is a standard integral, which evaluates to $-\ln|u|$. Substituting back $u = 2 - e^y$, we get $-\ln|2 - e^y| + C_1$.

Step 6: Evaluate the right integral

The right integral $\int \frac{1}{x+1} dx$ is also a standard integral. It evaluates to $\ln|x+1| + C_2$. We'll combine the constants of integration later.

Step 7: Combine the integrals

Equating the results of both integrals, we get $-\ln|2 - e^y| = \ln|x+1| + C$, where $C = C_2 - C_1$ is the combined constant of integration. This is the general solution to the differential equation.

Step 8: Apply initial condition

We are given the initial condition $y=0$ when $x=0$. Substitute these values into the general solution to find the specific value of the constant $C$. Since $e^0 = 1$ and $\ln(1) = 0$, we find that $C=0$.

Step 9: Substitute C back into the solution

Now that we have determined $C=0$, we substitute this value back into the general solution to obtain the particular solution. This equation implicitly defines $y$ in terms of $x$.

Step 10: Simplify and solve for y (optional)

To express $y$ explicitly, we can perform algebraic manipulations. We move the negative sign into the logarithm using the property $a\ln b = \ln b^a$. Then, we equate the arguments of the logarithms and solve for $e^y$. Taking the natural logarithm of both sides gives us $y$ as a function of $x$.

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