The solution of the differential equation 3 e^x ydx + (1 - e^x ) ^2 ydy = 0 is

The solution of the differential equation 3 e^x ydx + (1 - e^x ) …

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MCQ

The solution of the differential equation $3{e^x}\tan ydx + (1 - {e^x}){\sec ^2}ydy = 0$ is

✓
$\tan y = c{(1 - {e^x})^3}$
B
${(1 - {e^x})^3}\tan y = c$
C
$\tan y = c(1 - {e^x})$
D
$(1 - {e^x})\tan y = c$

✓

Answer

Correct option: A.

$\tan y = c{(1 - {e^x})^3}$

a
(a) It can be written in the form of

$\frac{{{{\sec }^2}y}}{{\tan y}}dy = - 3\frac{{{e^x}}}{{1 - {e^x}}}dx$

$\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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