Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y)$ and $Q(y)$ are functions of $y$.
• Integrating Factor (IF): The integrating factor is given by $IF = e^{\int P(y) \, dy}$.
• General Solution: The general solution of the first-order linear differential equation is $x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C$, where $C$ is the constant of integration.

Step-by-Step Solution

Step 1: Transform the Equation into Standard Linear Form

The given differential equation is $\frac{dx}{dy} = \frac{1+x-y^2}{y}$. We want to rewrite this in the standard form $\frac{dx}{dy} + P(y)x = Q(y)$.

First, we rewrite the right-hand side: $\frac{dx}{dy} = \frac{1}{y} + \frac{x}{y} - \frac{y^2}{y} = \frac{1}{y} + \frac{x}{y} - y$

Next, move the term containing $x$ to the left side: $\frac{dx}{dy} - \frac{x}{y} = \frac{1}{y} - y$

Rewrite this as: $\frac{dx}{dy} + \left(-\frac{1}{y}\right)x = \frac{1-y^2}{y}$

Comparing with the standard form, we identify: $P(y) = -\frac{1}{y}$ $Q(y) = \frac{1-y^2}{y}$

Step 2: Calculate the Integrating Factor (IF)

The integrating factor is given by $IF = e^{\int P(y) \, dy}$. Substitute $P(y)$: $IF = e^{\int -\frac{1}{y} \, dy}$

Now, integrate: $\int -\frac{1}{y} \, dy = -\ln|y|$

Therefore, the integrating factor is: $IF = e^{-\ln|y|} = e^{\ln|y|^{-1}} = |y|^{-1} = \frac{1}{|y|}$

Since the initial condition is $x(1)=1$, we can assume $y>0$ in the region of interest, so $|y| = y$. $IF = \frac{1}{y}$

Step 3: Find the General Solution

The general solution is given by $x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C$. Substitute $IF = \frac{1}{y}$ and $Q(y) = \frac{1-y^2}{y}$: $x \cdot \left(\frac{1}{y}\right) = \int \left(\frac{1-y^2}{y}\right) \cdot \left(\frac{1}{y}\right) \, dy + C$ $\frac{x}{y} = \int \frac{1-y^2}{y^2} \, dy + C$ $\frac{x}{y} = \int \left(\frac{1}{y^2} - 1\right) \, dy + C$ $\frac{x}{y} = \int (y^{-2} - 1) \, dy + C$ Now, integrate: $\frac{x}{y} = \frac{y^{-1}}{-1} - y + C$ $\frac{x}{y} = -\frac{1}{y} - y + C$ Multiply by $y$ to express $x$ as a function of $y$: $x = -1 - y^2 + Cy$

Step 4: Apply the Initial Condition

We are given $x(1) = 1$. Substitute $x=1$ and $y=1$ into the general solution: $1 = -1 - (1)^2 + C(1)$ $1 = -1 - 1 + C$ $1 = -2 + C$ $C = 3$

Step 5: Formulate the Particular Solution

Substitute $C=3$ into the general solution: $x = -1 - y^2 + 3y$

Step 6: Evaluate $5x(2)$

First, find $x(2)$ by substituting $y=2$ into the particular solution: $x(2) = -1 - (2)^2 + 3(2)$ $x(2) = -1 - 4 + 6$ $x(2) = 1$

Now, calculate $5x(2)$: $5x(2) = 5 \times 1 = 5$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when rearranging the equation and integrating.
• Integrating Factor: Ensure you correctly calculate the integrating factor. A mistake here will invalidate the rest of the solution.
• Constant of Integration: Don't forget the constant of integration, $C$, when finding the general solution.

Summary

We solved the first-order linear differential equation by transforming it into standard form, finding the integrating factor, determining the general solution, applying the initial condition to find the particular solution, and finally evaluating $5x(2)$.

The final answer is \boxed{5}.