MCQ
વિકલ સમીકરણ $3{e^x}\tan ydx + (1 - {e^x}){\sec ^2}ydy = 0$ નો ઉકેલ મેળવો.
- ✓$\tan y = c{(1 - {e^x})^3}$
- B${(1 - {e^x})^3}\tan y = c$
- C$\tan y = c(1 - {e^x})$
- D$(1 - {e^x})\tan y = c$
$\frac{{{{\sec }^2}y}}{{\tan y}}dy = - 3\frac{{{e^x}}}{{1 - {e^x}}}dx$
$\int {\frac{{{{\sec }^2}y}}{{\tan y}}} dy = - 3\int {\frac{{{e^x}}}{{1 - {e^x}}}dx} $
==> $\log (\tan y) = 3\log (1 - {e^x}) + \log c$ ==> $\tan y = c{(1 - {e^x})^3}$.
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