JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y(x)\) be the solution of the differential equation \(2 x^{2} d y+\left(e^{y}-2 x\right) d x=0, x>0\). If \(y(e)=1\), then \(\mathrm{y}(1)\) is equal to :
- A \(0\)
- B \(2\)
- C \(\log _{\mathrm{e}} 2\)
- D \(\log _{c}(2 \mathrm{e})\)
Answer & Solution
Correct Answer
(C) \(\log _{\mathrm{e}} 2\)
Step-by-step Solution
Detailed explanation
\(2 x^{2} d y+\left(e^{y}-2 x\right) d x=0\) \(\frac{d y}{d x}+\frac{e^{y}-2 x}{2 x^{2}}=0 \Rightarrow \frac{d y}{d x}+\frac{e^{y}}{2 x^{2}}-\frac{1}{x}=0\) \(e^{-y} \frac{d y}{d x}-\frac{e^{-y}}{x}=-\frac{1}{2 x^{2}} \Rightarrow \text { Put } e^{-y}=z\)…
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