JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(\frac{d y}{d x}=1+x e^{y-x},-\sqrt{2}\,<\,x\,<\,\sqrt{2}, y (0)=0\) then, the minimum value of \(y(x)\) , \(\mathrm{x} \in(-\sqrt{2}, \sqrt{2})\) is equal to:
- A \((1-\sqrt{3})-\log _{e}(\sqrt{3}-1)\)
- B \((2+\sqrt{3})+\log _{e} 2\)
- C \((2-\sqrt{3})-\log _{e} 2\)
- D \((1+\sqrt{3})-\log _{e}(\sqrt{3}-1)\)
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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