Question

If \(y=y(x)\) is the solution curve of the differential equation
\(x^2\,dy + \left(y - \frac{1}{x}\right)dx = 0\)
; \(x \gt 0\) and \(y(1)=1\), then \(y\left(\frac{1}{2}\right)\) is equal to :

• A. \(\frac{3}{2} - \frac{1}{\sqrt{e}}\)
• B. \(3 + \frac{1}{\sqrt{e}}\)
• C. \(3 + e\)
• D. \(3 - e\)

Hint:

Rewrite as the linear equation \(\frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3}\) and use the integrating factor \(e^{-1/x}\).

Answer 1

The Correct Option is D

Divide by \(x^2\,dx\): \(\frac{dy}{dx}+\frac{1}{x^2}\,y=\frac{1}{x^3}\).
Integrating factor \(=e^{\int x^{-2}dx}=e^{-1/x}\).
\(y\,e^{-1/x}=\int \frac{e^{-1/x}}{x^3}dx\). Put \(t=\frac{1}{x}\), \(dx=-\frac{dt}{t^2}\): integral \(=-\int t\,e^{-t}dt=(t+1)e^{-t}+C\).
So \(y\,e^{-1/x}=\left(\frac{1}{x}+1\right)e^{-1/x}+C\), i.e. \(y=1+\frac{1}{x}+Ce^{1/x}\).
\(y(1)=1\): \(1=2+Ce\), so \(C=-\frac{1}{e}\) and \(y=1+\frac{1}{x}-e^{\frac{1}{x}-1}\).
At \(x=\frac12\): \(y=1+2-e^{2-1}=3-e\).
Answer: (4) \(3 - e\)