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JEE Mains · Maths · STD 12 - 9. differential equations

मान लीजिए कि अवकल समीकरण \(y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0\) का हल \(x=x(y)\) है। यदि \(x(1)=1\), तो \(x\left(\frac{1}{2}\right)\) क्या है?

  1. A \(\frac{1}{2}+\mathrm{e}\)
  2. B \(3+e\)
  3. C \(3-e\)
  4. D \(\frac{3}{2}+e\)
Verified Solution

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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