Key Concepts and Formulas

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

Step-by-Step Solution

Step 1: Rewrite the given differential equation in the standard form.

We are given the differential equation ${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$. We need to rearrange this into the form $\frac{dy}{dx} + P(x)y = Q(x)$.

Divide the entire equation by $x^2 dx$: $\frac{x^2 dy}{x^2 dx} + \frac{(y - \frac{1}{x})dx}{x^2 dx} = 0$ $\frac{dy}{dx} + \frac{y}{x^2} - \frac{1}{x^3} = 0$ $\frac{dy}{dx} + \frac{1}{x^2}y = \frac{1}{x^3}$

Now, we have the equation in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{1}{x^2}$ and $Q(x) = \frac{1}{x^3}$.

Step 2: Calculate the Integrating Factor (IF).

The integrating factor is given by $IF = e^{\int P(x) dx}$. In our case, $P(x) = \frac{1}{x^2}$. $IF = e^{\int \frac{1}{x^2} dx} = e^{\int x^{-2} dx} = e^{\frac{x^{-1}}{-1}} = e^{-\frac{1}{x}}$

Step 3: Find the general solution of the differential equation.

The general solution is given by $y(x) \cdot IF = \int Q(x) \cdot IF \, dx + C$. We have $IF = e^{-\frac{1}{x}}$ and $Q(x) = \frac{1}{x^3}$. $y \cdot e^{-\frac{1}{x}} = \int \frac{1}{x^3} \cdot e^{-\frac{1}{x}} dx + C$

Let $u = -\frac{1}{x}$, so $du = \frac{1}{x^2} dx$. Then, $x^2 = \frac{1}{u^2}$, so $x^3 = \frac{1}{-u^3}$. Thus, $dx = x^2 du = \frac{1}{u^2}du$. Substituting into the integral: $\int \frac{1}{x^3} e^{-\frac{1}{x}} dx = \int (-u^3) e^u \frac{1}{u^2} du = \int -u e^u du = - \int u e^u du$

Using integration by parts, let $v = u$ and $dw = e^u du$. Then $dv = du$ and $w = e^u$. So $\int u e^u du = u e^u - \int e^u du = u e^u - e^u$.

Therefore, $\int \frac{1}{x^3} e^{-\frac{1}{x}} dx = - (u e^u - e^u) = e^u - ue^u = e^{-\frac{1}{x}} - (-\frac{1}{x})e^{-\frac{1}{x}} = e^{-\frac{1}{x}} + \frac{1}{x} e^{-\frac{1}{x}}$.

Substituting back into the general solution: $y e^{-\frac{1}{x}} = e^{-\frac{1}{x}} + \frac{1}{x} e^{-\frac{1}{x}} + C$ Divide by $e^{-\frac{1}{x}}$: $y = 1 + \frac{1}{x} + C e^{\frac{1}{x}}$

Step 4: Apply the initial condition y(1) = 1 to find the particular solution.

We are given that $y(1) = 1$. Substituting $x = 1$ and $y = 1$ into the general solution: $1 = 1 + \frac{1}{1} + C e^{\frac{1}{1}}$ $1 = 1 + 1 + C e$ $-1 = C e$ $C = -\frac{1}{e}$

Thus, the particular solution is: $y = 1 + \frac{1}{x} - \frac{1}{e} e^{\frac{1}{x}} = 1 + \frac{1}{x} - e^{\frac{1}{x} - 1}$

Step 5: Evaluate the solution at x = 1/2.

We need to find $y\left(\frac{1}{2}\right)$. $y\left(\frac{1}{2}\right) = 1 + \frac{1}{\frac{1}{2}} - e^{\frac{1}{\frac{1}{2}} - 1} = 1 + 2 - e^{2 - 1} = 3 - e$

Step 6: Re-examine the integration by parts to correct the error.

There was an error in the previous integration. It should have been: $y \cdot e^{-\frac{1}{x}} = \int \frac{1}{x^3} \cdot e^{-\frac{1}{x}} dx + C$ Let $u = -\frac{1}{x}$. Then $du = \frac{1}{x^2} dx$. So $\int \frac{1}{x^3}e^{-\frac{1}{x}}dx = \int \frac{1}{x} \cdot \frac{1}{x^2} e^{-\frac{1}{x}} dx = \int -u e^u du$. Integrating by parts, let $v = -u$ and $dw = e^u du$, so $dv = -du$ and $w = e^u$. $\int -u e^u du = -u e^u - \int -e^u du = -ue^u + e^u + C_1$. Substituting back, we get $\frac{1}{x}e^{-\frac{1}{x}} + e^{-\frac{1}{x}} + C_1$ So $y e^{-\frac{1}{x}} = \frac{1}{x}e^{-\frac{1}{x}} + e^{-\frac{1}{x}} + C$. Dividing by $e^{-\frac{1}{x}}$, we get $y = \frac{1}{x} + 1 + Ce^{\frac{1}{x}}$. Using $y(1) = 1$, we have $1 = 1 + 1 + Ce^1$, which gives $C = -\frac{1}{e}$. Therefore $y = \frac{1}{x} + 1 - \frac{1}{e}e^{\frac{1}{x}}$. When $x = \frac{1}{2}$, $y(\frac{1}{2}) = 2 + 1 - \frac{1}{e}e^2 = 3 - e$. This is still incorrect.

Step 7: Re-re-examine the integration by parts. Let $u = -1/x$, then $du = (1/x^2) dx$. We have $\int \frac{1}{x^3} e^{-1/x} dx = \int \frac{1}{x} \frac{1}{x^2} e^{-1/x} dx = \int -u e^u du$ Using integration by parts, let $v = -u$ and $dw = e^u du$. Then $dv = -du$ and $w = e^u$. So $\int -u e^u du = -u e^u - \int -e^u du = -u e^u + e^u + C_1 = e^u(1-u) + C_1 = e^{-1/x} (1 + \frac{1}{x}) + C_1$. Therefore, $ye^{-1/x} = e^{-1/x}(1 + \frac{1}{x}) + C$. Dividing by $e^{-1/x}$, we have $y = 1 + \frac{1}{x} + Ce^{1/x}$. Given $y(1) = 1$, we have $1 = 1 + 1 + Ce$, so $C = -1/e$. Thus $y = 1 + \frac{1}{x} - \frac{1}{e} e^{1/x}$.

We want $y(1/2)$. $y(1/2) = 1 + 2 - \frac{1}{e} e^2 = 3 - e$. This is still incorrect, unfortunately.

Step 8: Double-Check the Given Solution

Let's take a derivative of the supposed solution to see if it fits the equation. $y = 1 + \frac{1}{x} - \frac{1}{e} e^{1/x}$ $dy/dx = -1/x^2 - \frac{1}{e} e^{1/x} \cdot (-1/x^2) = -1/x^2 + \frac{1}{ex^2} e^{1/x}$. Substituting into the original equation: $x^2 dy + (y - 1/x)dx = 0$ $x^2 (-1/x^2 + \frac{1}{ex^2} e^{1/x}) dx + (1 + 1/x - \frac{1}{e} e^{1/x} - 1/x) dx = 0$ $(-1 + \frac{1}{e} e^{1/x}) dx + (1 - \frac{1}{e} e^{1/x}) dx = 0$. $-dx + \frac{1}{e} e^{1/x}dx + dx - \frac{1}{e} e^{1/x}dx = 0$. So the solution is correct.

We want $y(1/2)$. $y(1/2) = 1 + 2 - \frac{1}{e} e^2 = 3 - e$.

Let us re-examine the problem statement. The equation is ${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$. Then $x^2 dy + ydx = \frac{1}{x} dx$. Divide by $x^2$, $dy + \frac{y}{x^2} dx = \frac{1}{x^3} dx$. The integrating factor is $e^{\int \frac{1}{x^2} dx} = e^{-1/x}$. The solution is $ye^{-1/x} = \int \frac{1}{x^3} e^{-1/x} dx + C$. Let $u = -1/x$. Then $du = \frac{1}{x^2} dx$. So $\int \frac{1}{x^3} e^{-1/x} dx = \int \frac{1}{x} \frac{1}{x^2} e^{-1/x} dx = \int -u e^u du = -ue^u + e^u = \frac{1}{x} e^{-1/x} + e^{-1/x}$. So $ye^{-1/x} = \frac{1}{x} e^{-1/x} + e^{-1/x} + C$. Then $y = \frac{1}{x} + 1 + C e^{1/x}$. $y(1) = 1 = 1 + 1 + Ce$, so $C = -1/e$. Then $y = \frac{1}{x} + 1 - \frac{1}{e} e^{1/x}$. When $x = 1/2$, $y = 2 + 1 - \frac{1}{e} e^2 = 3 - e$.

Still $3-e$, but that is not any of the provided answers.

If instead, we had $y = 1+1/x - Ce^{1/x}$, and $C = 1/\sqrt{e}$, then we would get, for $x=1/2$, $y = 3 - \frac{1}{\sqrt{e}} e^2 = 3 - e^{3/2}$.

Let's try to get option A. We need $3/2 - 1/\sqrt{e}$. If we had $y = 1+ 1/x + Ce^{1/x}$, and $y(1) = 1$, then $1 = 2 + Ce$, so $C = -1/e$. So $y = 1+ 1/x - \frac{1}{e} e^{1/x}$. Then $y(1/2) = 3 - \frac{1}{e} e^2 = 3 - e$.

Let's try manipulating the answer. $\frac{3}{2} - \frac{1}{\sqrt{e}} = 3 - \frac{3}{2} - \frac{1}{\sqrt{e}} = 3 - (\frac{3}{2} + \frac{1}{\sqrt{e}})$.

If the original equation was ${x^2}dy + \left( {y + {1 \over x}} \right)dx = 0$, then $dy/dx + y/x^2 = -1/x^3$. Following the steps, $y = 1/x + 1 - (2+e^{-1/2}) e^{1/x}$.

Step 9: Correcting the Initial Condition

Let's assume that the initial condition was $y(1) = 2$ rather than 1. Then: $2 = 1 + \frac{1}{1} + Ce$ $2 = 1 + 1 + Ce$ $0 = Ce$ So $C = 0$. Then $y = 1 + \frac{1}{x}$. So $y(1/2) = 1 + 2 = 3$.

Step 10: Working Backwards from the Correct Answer Let's assume the answer is correct. Then $y(1/2) = \frac{3}{2} - \frac{1}{\sqrt{e}}$. Then $y(x) = 1 + \frac{1}{x} + Ce^{1/x}$. So $y(1/2) = 3 - Ce^2 = \frac{3}{2} - \frac{1}{\sqrt{e}}$. $Ce^2 = 3 - \frac{3}{2} + \frac{1}{\sqrt{e}} = \frac{3}{2} + \frac{1}{\sqrt{e}}$. So $C = (\frac{3}{2} + \frac{1}{\sqrt{e}}) e^{-2}$.

Then $y = 1 + \frac{1}{x} + (\frac{3}{2} + \frac{1}{\sqrt{e}}) e^{-2} e^{1/x}$.

This is still incorrect.

Step 11: Final Attempt - Re-examining the Integrating Factor Approach Starting again from: $x^2 dy + (y - \frac{1}{x}) dx = 0$ $dy + (\frac{y}{x^2} - \frac{1}{x^3}) dx = 0$ $\frac{dy}{dx} + \frac{y}{x^2} = \frac{1}{x^3}$ Integrating factor is $e^{\int \frac{1}{x^2} dx} = e^{-1/x}$. So $y e^{-1/x} = \int \frac{1}{x^3} e^{-1/x} dx + C$. Let $u = -1/x$. Then $du = dx/x^2$. So $\int \frac{1}{x^3} e^{-1/x} dx = \int \frac{1}{x} e^u du dx/dx^2 = \int -ue^u du$. $\int -ue^u du = -ue^u + e^u + C = (1-u)e^u + C = (1 + \frac{1}{x}) e^{-1/x} + C$. So $ye^{-1/x} = (1 + \frac{1}{x}) e^{-1/x} + C$. $y = 1 + \frac{1}{x} + Ce^{1/x}$. Now $y(1) = 1$. So $1 = 1 + 1 + Ce$. $C = -1/e$. So $y = 1 + \frac{1}{x} - \frac{1}{e} e^{1/x}$. $y(1/2) = 1 + 2 - \frac{1}{e} e^2 = 3 - e$.

Since we are certain about the derivation now, and the answer must be (A), there must be an error in the question or answer key. Given the options, there are no typos that can convert 3-e into the correct answer.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating the integrating factor and performing integration by parts. A single sign error can throw off the entire solution.
• Integration by Parts: Remember the formula for integration by parts: $\int u \, dv = uv - \int v \, du$. Choose $u$ and $dv$ carefully to simplify the integral.
• Checking the Solution: After finding the particular solution, it's a good practice to differentiate it and substitute it back into the original differential equation to verify that it satisfies the equation.

Summary

We aimed to solve the first-order linear differential equation ${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$ with the initial condition $y(1) = 1$. We found the general solution using the integrating factor method and then applied the initial condition to find the particular solution. We then evaluated this solution at $x = \frac{1}{2}$. After multiple careful attempts, the derived answer consistently comes out to $3 - e$, which does not match the provided correct answer.

Final Answer

The final answer is \boxed{3 - e}. However, this does not match any of the provided options. There may be an error in the provided options or the correct answer.