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Step-by-Step Solution
Step 1: Rewrite the given differential equation
We are given
$ x^{3}\,dy + (x\,y - 1)\,dx = 0,\quad x > 0. $
Divide both sides by $dx$ (assuming $dx \neq 0$) to get
$ x^{3}\,\frac{dy}{dx} + x\,y - 1 = 0. $
Rearrange to isolate $dy/dx$:
$ x^{3}\,\frac{dy}{dx} = 1 - x\,y. $
Hence,
$ \frac{dy}{dx} + \frac{y}{x^{2}} = \frac{1}{x^{3}}. $
Step 2: Identify the integrating factor
We have the linear differential equation in the form
$ \frac{dy}{dx} + P(x)\,y = Q(x), $
where $P(x) = \frac{1}{x^{2}}$ and $Q(x) = \frac{1}{x^{3}}.$
The integrating factor (I.F.) is given by
$ \text{I.F.} = e^{\int P(x)\,dx} = e^{\int \frac{1}{x^2}\,dx} = e^{-\frac{1}{x}}. $
Step 3: Multiply through by the integrating factor
Multiply the entire differential equation by $e^{-\frac{1}{x}}$ to get
$ e^{-\frac{1}{x}}\frac{dy}{dx} + \frac{y}{x^2}\,e^{-\frac{1}{x}} = \frac{1}{x^3}\,e^{-\frac{1}{x}}. $
Observe that the left-hand side is the derivative of $y\,e^{-\frac{1}{x}}$ with respect to $x$:
$ \frac{d}{dx}\bigl(y\,e^{-\frac{1}{x}}\bigr) = \frac{1}{x^3}\,e^{-\frac{1}{x}}. $
Step 4: Integrate both sides
We now integrate with respect to $x$:
$ y\,e^{-\frac{1}{x}} = \int \frac{e^{-\frac{1}{x}}}{x^3}\,dx + C, $
where $C$ is the constant of integration.
Step 5: Perform the substitution to evaluate the integral
Let $ t = -\frac{1}{x}. $ Then
$ dt = \frac{1}{x^2}\,dx \quad \Longrightarrow\quad dx = x^2\,dt. $
Also note that with $ t = -\frac{1}{x}, $ we have
$ \frac{e^{-\frac{1}{x}}}{x^3}\,dx = e^t \left(\frac{1}{x^3}\right)\left(x^2\,dt\right) = \frac{e^t}{x}\,dt. $
Since $t = -\frac{1}{x}$, we get $ x = -\frac{1}{t}. $ Hence
$ \frac{e^t}{x} = \frac{e^t}{-\frac{1}{t}} = -\,t\,e^t. $
Therefore,
$ \int \frac{e^{-\frac{1}{x}}}{x^3}\,dx
= \int -t\,e^t\,dt. $
Step 6: Integrate in terms of t
We use integration by parts or a known result:
$ \int t\,e^t\,dt = t\,e^t - e^t + \text{constant}. $
Hence,
$ \int -t\,e^t\,dt
= -\bigl(t\,e^t - e^t\bigr) + \text{constant}
= -t\,e^t + e^t + \text{constant}. $
Substituting back, we write the general solution as
$ y\,e^{-\frac{1}{x}} = -\bigl(t\,e^t - e^t\bigr) + C'
= e^t - t\,e^t + C'. $
Since $t = -\frac{1}{x}$ and $e^t = e^{-\frac{1}{x}}$, we get
$ y\,e^{-\frac{1}{x}} = e^{-\frac{1}{x}} - \left(-\frac{1}{x}\right)e^{-\frac{1}{x}} + C'
= e^{-\frac{1}{x}} + \frac{e^{-\frac{1}{x}}}{x} + C'. $
Let us denote the constant by $C$ for simplicity:
$ y\,e^{-\frac{1}{x}} = e^{-\frac{1}{x}} + \frac{e^{-\frac{1}{x}}}{x} + C. $
Step 7: Apply the initial condition y(1/2) = 3 − e
When $x = \tfrac{1}{2},\; y\left(\tfrac{1}{2}\right) = 3 - e.$ Substitute in the general solution:
$ \bigl(3 - e\bigr)\,e^{-2}
= e^{-2} + \frac{e^{-2}}{\tfrac{1}{2}} + C. $
Observe that $\frac{1}{\frac{1}{2}} = 2,$ so the right-hand side becomes
$ e^{-2} + 2\,e^{-2} + C = 3e^{-2} + C. $
Thus,
$ (3 - e)\,e^{-2} = 3\,e^{-2} + C \quad\Longrightarrow\quad C = (3 - e)\,e^{-2} - 3\,e^{-2}. $
Factor out $e^{-2}$:
$ C = e^{-2}\bigl(3 - e - 3\bigr) = e^{-2}\bigl(- e\bigr) = -\,\frac{1}{e}. $
Step 8: Find y(1)
Now substitute $x = 1$ and $C = -\tfrac{1}{e}$ back into
$ y\,e^{-\frac{1}{x}} = e^{-\frac{1}{x}} + \frac{e^{-\frac{1}{x}}}{x} + C. $
For $x=1$, we have $e^{-\frac{1}{1}} = e^{-1}$, and
$ y(1)\,e^{-1} = e^{-1} + \frac{e^{-1}}{1} - \frac{1}{e}. $
Hence,
$ y(1)\,e^{-1} = e^{-1} + e^{-1} - e^{-1}
= e^{-1}. $
Divide both sides by $e^{-1}$:
$ y(1) = 1. $
Final Answer
Therefore, the value of $y(1)$ is $1$.
