MCQ
જો $y=y(x)$ એ વિકલ સમીકરણ $d y=e^{a x+y} d x ; \alpha \in N$ નો ઉકેલ છે અને જો $y\left(\log _{e} 2\right)=\log _{e} 2$ અને $y(0)=\log _{e}\left(\frac{1}{2}\right)$, હોય તો $\alpha$ ની કિમંત મેળવો.
- A$1$
- B$2$
- C$3$
- D$5$
$\Rightarrow-\mathrm{e}^{-y}=\frac{\mathrm{e}^{\alpha \mathrm{x}}}{\alpha}+\mathrm{c}....(i)$
Put $(x, y)=(\ell n 2, \ell n 2)$
$\frac{-1}{2}=\frac{2^{\alpha}}{\alpha}+C....(ii)$
Put $(x, y) \equiv(0,-\ell n 2)$ in $(i)$
$-2=\frac{1}{\alpha}+C....(iii)$
$(ii) - (iii)$
$\frac{2 \alpha-1}{\alpha}=\frac{3}{2}$
$\Rightarrow \alpha=2(a s \alpha \in N)$
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