JEE Mains · Maths · STD 12 - 9. differential equations
Let \(x=x(y)\) be the solution of the differential equation \(y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0\). If \(x(1)=1\), then \(x\left(\frac{1}{2}\right)\) is :
- A \(\frac{1}{2}+\mathrm{e}\)
- B \(3+e\)
- C \(3-e\)
- D \(\frac{3}{2}+e\)
Answer & Solution
Correct Answer
(C) \(3-e\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & y^2 d x+\left(x-\frac{1}{y}\right) d y=0 \\ & y^2 d x=\left(\frac{1}{y}-x\right) d y \\ & \Rightarrow y^2 \frac{d x}{d y}=\frac{1}{y}-x \\ & \Rightarrow \frac{d x}{d y}+\frac{x}{y^2}=\frac{1}{y^3} \\ & \text { I.F. }=e^{\int \frac{1}{y^2} d y}=e^{\frac{-1}{y}}…
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