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

माना अवकल समीकरण, \(\frac{ dy }{ dx }=1+ xe ^{ y - x },-\sqrt{2}< x <\sqrt{2}, y (0)=0\) का हल \(y = y ( x )\) है। तो \(x \in(-\sqrt{2}, \sqrt{2})\) में \(y ( x )\) का न्यूनतम मान बराबर है

  1. A \((1-\sqrt{3})-\log _{e}(\sqrt{3}-1)\)
  2. B \((2+\sqrt{3})+\log _{e} 2\)
  3. C \((2-\sqrt{3})-\log _{e} 2\)
  4. D \((1+\sqrt{3})-\log _{e}(\sqrt{3}-1)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \((1-\sqrt{3})-\log _{e}(\sqrt{3}-1)\)

Step-by-step Solution

Detailed explanation

\(\frac{d y-d x}{e^{y-x}}=x \,d x\) \(\Rightarrow \frac{d y-d x}{e^{y-x}}=x \,d x\) \(\Rightarrow-e^{x-y}=\frac{x^{2}}{2}+c\) At \(x=0, y=0 \Rightarrow c=-1\) \(\Rightarrow e^{x-y}=\frac{2-x^{2}}{2}\) \(\Rightarrow y=x-\ell n\left(\frac{2-x^{2}}{2}\right)\)…
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