Question:
Consider the differential equation, \(y^2\,dx + \left(x - \frac{1}{y}\right)dy = 0\), If value of y is 1 when x = 1, then the value of x for which y = 2, is :
Consider the differential equation, \(y^2\,dx + \left(x - \frac{1}{y}\right)dy = 0\), If value of y is 1 when x = 1, then the value of x for which y = 2, is :
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Treat x as a function of y: \(\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}\), a linear equation with integrating factor \(e^{-1/y}\).
Updated On: Nov 5, 2026
- \(\frac{3}{2} - \frac{1}{\sqrt{e}}\)
- \(\frac{1}{2} + \frac{1}{\sqrt{e}}\)
- \(\frac{5}{2} + \frac{1}{\sqrt{e}}\)
- \(\frac{3}{2} - \sqrt{e}\)
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The Correct Option is A
Solution and Explanation
\(y^2\frac{dx}{dy} + x - \frac1y = 0 \Rightarrow \frac{dx}{dy} + \frac{1}{y^2}x = \frac{1}{y^3}\).
I.F. \(= e^{\int dy/y^2} = e^{-1/y}\).
\(x e^{-1/y} = \int \frac{e^{-1/y}}{y^3}dy\). Put \(t=1/y\), \(dt=-dy/y^2\): the integral becomes \(-\int t e^{-t}dt = (t+1)e^{-t} + C\).
So \(x e^{-1/y} = \left(\frac1y + 1\right)e^{-1/y} + C\).
At \(x=1,y=1\): \(e^{-1} = 2e^{-1} + C \Rightarrow C = -e^{-1}\).
At \(y=2\): \(x e^{-1/2} = \frac32 e^{-1/2} - e^{-1} \Rightarrow x = \frac32 - e^{-1/2} = \frac32 - \frac{1}{\sqrt e}\).
Answer: (1) \(\frac{3}{2} - \frac{1}{\sqrt{e}}\)
I.F. \(= e^{\int dy/y^2} = e^{-1/y}\).
\(x e^{-1/y} = \int \frac{e^{-1/y}}{y^3}dy\). Put \(t=1/y\), \(dt=-dy/y^2\): the integral becomes \(-\int t e^{-t}dt = (t+1)e^{-t} + C\).
So \(x e^{-1/y} = \left(\frac1y + 1\right)e^{-1/y} + C\).
At \(x=1,y=1\): \(e^{-1} = 2e^{-1} + C \Rightarrow C = -e^{-1}\).
At \(y=2\): \(x e^{-1/2} = \frac32 e^{-1/2} - e^{-1} \Rightarrow x = \frac32 - e^{-1/2} = \frac32 - \frac{1}{\sqrt e}\).
Answer: (1) \(\frac{3}{2} - \frac{1}{\sqrt{e}}\)
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