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Step-by-Step Solution
1. Rewrite the Differential Equation in Standard Form
The given differential equation is
$ x^2\,dy + \bigl(y - \tfrac{1}{x}\bigr)\,dx = 0 $ for $ x > 0 $.
We rearrange terms to separate $ dy $ and $ dx $:
$ x^2 \,\frac{dy}{dx} + y - \frac{1}{x} = 0 \quad\Longrightarrow\quad x^2\,\frac{dy}{dx} = \frac{1}{x} - y. $
Thus, we obtain
$ \frac{dy}{dx} = \frac{1}{x^3} - \frac{y}{x^2}. $
2. Recognize the Form of a Linear Differential Equation
We can write this as:
$ \frac{dy}{dx} + \frac{1}{x^2}\,y = \frac{1}{x^3}, $
which is a linear first-order differential equation of the form
$ \frac{dy}{dx} + P(x)\,y = Q(x). $
Here,
$ P(x) = \frac{1}{x^2} $
and
$ Q(x) = \frac{1}{x^3}. $
3. Find the Integrating Factor
The integrating factor $ \mu(x) $ is given by
$ \mu(x) = e^{\int P(x)\,dx} = e^{\int \frac{1}{x^2}\,dx} = e^{-\frac{1}{x}}. $
4. Multiply Through by the Integrating Factor
Multiplying both sides of the differential equation by $ e^{-\frac{1}{x}} $ gives:
$ e^{-\frac{1}{x}} \frac{dy}{dx} + \frac{1}{x^2} e^{-\frac{1}{x}} y = \frac{1}{x^3} e^{-\frac{1}{x}}. $
The left-hand side becomes the derivative of
$ \bigl(y \,e^{-\frac{1}{x}}\bigr) $:
$ \frac{d}{dx}\Bigl(y \, e^{-\frac{1}{x}}\Bigr) = \frac{1}{x^3} \, e^{-\frac{1}{x}}. $
5. Integrate Both Sides
We integrate both sides with respect to $ x $:
$ y \, e^{-\frac{1}{x}} = \int \frac{1}{x^3} \, e^{-\frac{1}{x}} \, dx + C. $
To handle the integral
$ \int \frac{1}{x^3}\,e^{-\tfrac{1}{x}}\,dx, $
we use the substitution
$ u = -\tfrac{1}{x}, $
which leads to
$ \int \frac{1}{x^3}\,e^{-\tfrac{1}{x}}\,dx = e^{-\tfrac{1}{x}} \Bigl(1 + \tfrac{1}{x}\Bigr) + \text{constant}. $
Thus,
$ y \, e^{-\frac{1}{x}} = e^{-\frac{1}{x}}\left(1 + \frac{1}{x}\right) + C. $
6. Apply the Initial Condition to Find the Constant
We use $ y(1) = 1 $ to determine $ C $. At $ x = 1 $:
$ y(1)\,e^{-\tfrac{1}{1}} = e^{-1}(1 + 1) + C. $
Since $ y(1) = 1 $, the left-hand side is $ 1 \cdot e^{-1} = e^{-1} $. Hence,
$ e^{-1} = e^{-1}\times 2 + C \quad\Longrightarrow\quad C = e^{-1} - 2e^{-1} = - e^{-1}. $
7. Write the General Solution
Substitute $ C = -\frac{1}{e} $ back into the general solution:
$ y\,e^{-\frac{1}{x}} = e^{-\frac{1}{x}}\left(1 + \frac{1}{x}\right) - \frac{1}{e}, $
which simplifies to
$ y = 1 + \frac{1}{x} - \frac{1}{e}\,e^{\frac{1}{x}}. $
8. Evaluate $ y \bigl(\tfrac{1}{2}\bigr) $
Now, we find $ y\bigl(\tfrac{1}{2}\bigr) $:
$ y\Bigl(\tfrac{1}{2}\Bigr) = 1 + 2 - \frac{1}{e} \, e^{2}. $
Observe that
$ -\frac{1}{e}\, e^{2} = -e. $
Hence,
$ y\Bigl(\tfrac{1}{2}\Bigr) = 3 - e. $
9. Final Answer
Therefore, the value of $ y\bigl(\tfrac{1}{2}\bigr) $ is
$ 3 - e. $
