JEE Mains · Maths · STD 12 - 9. differential equations
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
- A \(\frac{3}{2} - \sqrt e \)
- B \(\frac{1}{2} + \frac{1}{{\sqrt e }}\)
- C \(\frac{3}{2} - \frac{1}{{\sqrt e }}\)
- D \(\frac{5}{2} + \frac{1}{{\sqrt e }}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{2} - \frac{1}{{\sqrt e }}\)
Step-by-step Solution
Detailed explanation
\(y^{2} d x+x d y=\frac{d y}{y}\) \(\frac{d x}{d y}+\frac{x}{y^{2}}=\frac{1}{y^{3}}\) \({\rm{IF}} = {{\rm{e}}^{\int {\frac{1}{{{{\rm{y}}^2}}}} {\rm{dy}}}} = {{\rm{e}}^{\frac{1}{{\rm{y}}}}}\) \(e^{\frac{1}{y}} \cdot x=\int e^{\frac{1}{y}} \cdot \frac{1}{y^{3}} d y+C\)…
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