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 $y$ only.
• Integrating Factor (I.F.): For the equation $\frac{dx}{dy} + P(y)x = Q(y)$, the integrating factor is $e^{\int P(y) dy}$.
• General Solution: The general solution for the equation $\frac{dx}{dy} + P(y)x = Q(y)$ is $x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C$.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$.

Step-by-Step Solution

Step 1: Transform the given differential equation into the standard linear form

The given differential equation is: $y^2 dx + \left(x - \frac{1}{y}\right) dy = 0$ We want to express it in the form $\frac{dx}{dy} + P(y)x = Q(y)$.

1. Separate the $dx$ and $dy$ terms: Move terms with $dy$ to the right side: $y^2 dx = -\left(x - \frac{1}{y}\right) dy$ $y^2 dx = \left(\frac{1}{y} - x\right) dy$

2. Form the derivative $\frac{dx}{dy}$: Divide both sides by $dy$: $y^2 \frac{dx}{dy} = \frac{1}{y} - x$

3. Make the coefficient of $\frac{dx}{dy}$ equal to 1: Divide the entire equation by $y^2$: $\frac{dx}{dy} = \frac{1}{y^3} - \frac{x}{y^2}$

4. Rearrange into the standard form $\frac{dx}{dy} + P(y)x = Q(y)$: Move the term containing $x$ to the left side: $\frac{dx}{dy} + \frac{1}{y^2}x = \frac{1}{y^3}$ Comparing this to the standard form, we identify $P(y) = \frac{1}{y^2}$ and $Q(y) = \frac{1}{y^3}$.

Step 2: Calculate the Integrating Factor (I.F.)

The integrating factor is given by $\text{I.F.} = e^{\int P(y) dy}$.

1. Calculate the integral of $P(y)$: $\int P(y) dy = \int \frac{1}{y^2} dy = \int y^{-2} dy = \frac{y^{-1}}{-1} = -\frac{1}{y}$

2. Compute the Integrating Factor: $\text{I.F.} = e^{\int P(y) dy} = e^{-1/y}$

Step 3: Find the General Solution

The general solution is given by $x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C$.

1. Substitute the I.F. and $Q(y)$ into the general solution formula: $x \cdot e^{-1/y} = \int \frac{1}{y^3} \cdot e^{-1/y} dy + C$

2. Evaluate the integral $\int \frac{1}{y^3} e^{-1/y} dy$: Let $t = -\frac{1}{y}$, so $dt = \frac{1}{y^2} dy$. Then $y = -\frac{1}{t}$, and $\frac{1}{y} = -t$. The integral becomes: $\int \frac{1}{y^3} e^{-1/y} dy = \int \frac{1}{y} e^{-1/y} \frac{1}{y^2} dy = \int (-t) e^t dt = -\int t e^t dt$ Using integration by parts with $u = t$ and $dv = e^t dt$, we get $du = dt$ and $v = e^t$. Thus, $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t + C_1 = e^t(t - 1) + C_1$ So, $-\int t e^t dt = -e^t(t - 1) + C_2 = e^t(1 - t) + C_2$ Substituting $t = -\frac{1}{y}$ back, we have: $e^{-1/y}\left(1 + \frac{1}{y}\right) + C_2$

3. Write the complete general solution: $x \cdot e^{-1/y} = e^{-1/y}\left(1 + \frac{1}{y}\right) + C$

4. Isolate $x$: Divide both sides by $e^{-1/y}$: $x = 1 + \frac{1}{y} + C e^{1/y}$

Step 4: Apply the initial condition to find C

We are given that $y = 1$ when $x = 1$. Substituting these values into the general solution: $1 = 1 + \frac{1}{1} + C e^{1/1}$ $1 = 1 + 1 + Ce$ $1 = 2 + Ce$ $Ce = -1$ $C = -\frac{1}{e}$ So the particular solution is: $x = 1 + \frac{1}{y} - \frac{1}{e} e^{1/y} = 1 + \frac{1}{y} - e^{(1/y)-1}$

Step 5: Find x when y = 2

We need to find the value of $x$ when $y = 2$. Substituting $y = 2$ into the particular solution: $x = 1 + \frac{1}{2} - e^{(1/2) - 1} = 1 + \frac{1}{2} - e^{-1/2} = \frac{3}{2} - \frac{1}{\sqrt{e}}$

Common Mistakes & Tips

• Remember to correctly identify $P(y)$ and $Q(y)$ when using the integrating factor method.
• Be careful when performing integration by parts, especially with the signs.
• Don't forget to substitute back to the original variable after integration when using substitution.

Summary

We transformed the given differential equation into a first-order linear differential equation, calculated the integrating factor, found the general solution, applied the initial condition to find the particular solution, and finally, determined the value of $x$ when $y = 2$. The value of x when y=2 is $\frac{3}{2} - \frac{1}{\sqrt{e}}$.

Final Answer The final answer is $\boxed{\frac{3}{2} - \frac{1}{\sqrt{e}}}$, which corresponds to option (A).