JEE Mains · Maths · STD 12 - 9. differential equations
If \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) is the solution of the differential equation, \(\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-1\right)=\mathrm{e}^{\mathrm{x}}\) such that \(\mathrm{y}(0)=0,\) then \(\mathrm{y}(1)\) is equal to
- A \(2+\log _{e} 2\)
- B \(2e\)
- C \(\log _{e} 2\)
- D \(1+\log _{e} 2\)
Answer & Solution
Correct Answer
(D) \(1+\log _{e} 2\)
Step-by-step Solution
Detailed explanation
\(e^{y} \frac{d y}{d x}-e^{y}=e^{x},\) Let \(e^{y}=t\) \(\Rightarrow \mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\) \(\frac{d t}{d x}-t=e^{x}\) I.F. \(=e^{\int-d x}=e^{-x}\)…
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