Key Concepts and Formulas

• Separable Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x)g(y)$ is separable.
• Solving Separable Equations: Separate variables and integrate both sides: $\int \frac{dy}{g(y)} = \int f(x) dx$.
• Exponential Integral: $\int e^x dx = e^x + C$

Step-by-Step Solution

Step 1: Separate the variables

We are given the differential equation: $\frac{dy}{dx} = xy - 1 + x - y$ Rearrange the terms on the right-hand side to factor: $\frac{dy}{dx} = x(y+1) - (y+1)$ $\frac{dy}{dx} = (x-1)(y+1)$ Now, separate the variables by dividing both sides by $(y+1)$ and multiplying both sides by $dx$: $\frac{dy}{y+1} = (x-1) dx$ The equation is now separated.

Step 2: Integrate both sides

Integrate both sides of the equation with respect to their respective variables: $\int \frac{dy}{y+1} = \int (x-1) dx$ The integral on the left side is: $\int \frac{dy}{y+1} = \ln|y+1| + C_1$ The integral on the right side is: $\int (x-1) dx = \frac{x^2}{2} - x + C_2$ Combining these, we have: $\ln|y+1| = \frac{x^2}{2} - x + C$ where $C = C_2 - C_1$ is an arbitrary constant.

Step 3: Apply the initial condition

We are given the initial condition $y(0) = 0$. Substitute $x=0$ and $y=0$ into the general solution: $\ln|0+1| = \frac{0^2}{2} - 0 + C$ $\ln(1) = 0 + C$ $0 = C$ Thus, $C = 0$.

Step 4: Find the particular solution

Substitute $C=0$ back into the general solution: $\ln|y+1| = \frac{x^2}{2} - x$ Since we expect y(1) to be positive, we can drop the absolute value. Exponentiate both sides to solve for $y$: $e^{\ln(y+1)} = e^{\frac{x^2}{2} - x}$ $y+1 = e^{\frac{x^2}{2} - x}$ $y = e^{\frac{x^2}{2} - x} - 1$

Step 5: Evaluate y(1)

Substitute $x=1$ into the particular solution to find $y(1)$: $y(1) = e^{\frac{1^2}{2} - 1} - 1$ $y(1) = e^{\frac{1}{2} - 1} - 1$ $y(1) = e^{-\frac{1}{2}} - 1$

Step 6: Check the options The correct answer is therefore $y(1) = e^{-\frac{1}{2}} - 1$. However, the correct answer is given as y(1) = 1. Let's re-evaluate the solution.

Step 1: Re-examine the factorization. $\frac{dy}{dx} = xy - 1 + x - y = xy - y + x - 1 = y(x-1) + (x-1) = (y+1)(x-1)$ Okay, the factorization is correct.

Step 2: Re-examine the Integration $\int \frac{dy}{y+1} = \int (x-1) dx$ $ln|y+1| = \frac{x^2}{2} - x + C$

Step 3: Re-examine the initial condition y(0) = 0 $ln|0+1| = \frac{0}{2} - 0 + C$ $0 = C$

Step 4: Re-examine the solution for y(x) $ln|y+1| = \frac{x^2}{2} - x$ $y+1 = e^{\frac{x^2}{2} - x}$ $y = e^{\frac{x^2}{2} - x} - 1$

Step 5: Re-examine y(1) $y(1) = e^{\frac{1}{2} - 1} - 1 = e^{-\frac{1}{2}} - 1$

Okay, we have made no errors. However, the answer provided is y(1) = 1. Let's consider if the question actually had a typo. Suppose the equation was $\frac{dy}{dx} = 1 - x + y - xy$. This would yield $\frac{dy}{dx} = (1-x) + y(1-x) = (1-x)(1+y)$ $\int \frac{dy}{1+y} = \int (1-x) dx$ $ln|1+y| = x - \frac{x^2}{2} + C$ $y(0) = 0 \implies ln(1) = 0 - 0 + C \implies C = 0$ $ln|1+y| = x - \frac{x^2}{2}$ $y = e^{x - \frac{x^2}{2}} - 1$ $y(1) = e^{1 - \frac{1}{2}} - 1 = e^{\frac{1}{2}} - 1$

This corresponds to option D.

However, if the question was $\frac{dy}{dx} = y + x - xy - 1 = y(1-x) + (x-1) = (y-1)(1-x)$, then $\frac{dy}{y-1} = (1-x) dx$ $ln|y-1| = x - \frac{x^2}{2} + C$ y(0) = 0, so $ln|-1| = 0 - 0 + C$ $0 = C$ $ln|y-1| = x - \frac{x^2}{2}$ $y-1 = e^{x - \frac{x^2}{2}}$ $y = e^{x - \frac{x^2}{2}} + 1$ $y(1) = e^{1 - \frac{1}{2}} + 1 = e^{\frac{1}{2}} + 1$

Still not giving us the right answer.

Let's assume that y(1) = 1 is correct. $1 = e^{\frac{1}{2} - 1} - 1$ $2 = e^{-\frac{1}{2}}$ $ln(2) = -\frac{1}{2}$ Which is not true.

Let's assume that the differential equation is $\frac{dy}{dx} = xy - y + 1 - x$ $\frac{dy}{dx} = y(x-1) - (x-1) = (y-1)(x-1)$ $\int \frac{dy}{y-1} = \int (x-1) dx$ $ln(y-1) = \frac{x^2}{2} - x + C$ $y(0) = 0, ln(-1) = 0 + C$ This is impossible.

Common Mistakes & Tips

• Double-check your factorization and integration steps.
• Be careful with signs when separating variables.
• Remember to apply the initial condition to find the particular solution.

Summary

We separated the variables, integrated both sides, and applied the initial condition to solve the differential equation. The solution we obtained is $y = e^{\frac{x^2}{2} - x} - 1$. Evaluating at $x=1$ gives $y(1) = e^{-\frac{1}{2}} - 1$. Since none of the options match this result, and we have checked and re-checked the solution, we must conclude that the answer key given is incorrect. There appears to be an error in the question or answer key. If we assume y(1)=1, this contradicts our derived solution. Thus, we assume the correct differential equation is $\frac{dy}{dx} = 1 - x + y - xy$, which would result in $y(1) = e^{\frac{1}{2}} - 1$.

Final Answer

The final answer should be $y(1) = e^{-\frac{1}{2}} - 1$ for the given equation. The correct answer provided, $y(1) = 1$, is incorrect. Given the choices, there is no correct option.