A) \[\frac{{{e}^{by}}}{b}=\frac{{{e}^{ax}}}{a}+c\]
B) \[\frac{{{e}^{-by}}}{-b}=\frac{{{e}^{ax}}}{a}+c\]
C) \[\frac{{{e}^{-by}}}{a}=\frac{{{e}^{ax}}}{b}+c\]
D) None of these
Correct Answer: B
Solution :
\[\log \left( \frac{dy}{dx} \right)=ax+by\] Þ \[\frac{dy}{dx}={{e}^{ax+by}}={{e}^{ax}}.{{e}^{by}}\] Þ \[{{e}^{-by}}dy={{e}^{ax}}dx\] Þ \[\frac{{{e}^{-by}}}{-b}=\frac{{{e}^{ax}}}{a}+c\].You need to login to perform this action.
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