JEE Mains · Maths · STD 12 - 9. differential equations
If \(y=y(x)\) is the solution curve of the differential equation \(x^{2} d y+\left(y-\frac{1}{x}\right) d x=0 \quad ; x>0\) and \(\mathrm{y}(1)=1\), then \(\mathrm{y}\left(\frac{1}{2}\right)\) is equal to :
- A \(\frac{3}{2}-\frac{1}{\sqrt{\mathrm{e}}}\)
- B \(3+\frac{1}{\sqrt{\mathrm{e}}}\)
- C \(3+\mathrm{e}\)
- D \(3-\mathrm{e}\)
Answer & Solution
Correct Answer
(D) \(3-\mathrm{e}\)
Step-by-step Solution
Detailed explanation
\(x^{2} d y+\left(y-\frac{1}{x}\right) d x=0: x>0, y(1)=1\) \(x^{2} d y+\frac{(x y-1)}{x} d x=0\) \(x^{2} d y=\frac{(x y-1)}{x} d x\) \(\frac{d y}{d x}=\frac{1-x y}{x^{3}}\) \(\frac{d y}{d x}=\frac{1}{x^{3}}-\frac{y}{x^{2}}\)…
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