What is the solution of the differential equation \[\log \left( {\dfrac{{dy}}{{dx}}} \right) = ax + by\]?
A. \[\dfrac{{{e^{by}}}}{b} = \dfrac{{{e^{ax}}}}{a} + c\]
B. \[\dfrac{{{e^{ - by}}}}{{ - b}} = \dfrac{{{e^{ax}}}}{a} + c\]
C. \[\dfrac{{{e^{ - by}}}}{a} = \dfrac{{{e^{ax}}}}{b} + c\]
D. None of these

Hint: First we will apply the antilogarithm formula in the given differential equation. Then separate the variables of the differential equation and integrate both sides to get the solution.

Formula Used: Integrating formula:
\[\int {{e^{mx}}dx} = \dfrac{{{e^{mx}}}}{m} + c\]
Antilogarithm formula:
\[a = {\log _b}c\]
\[ \Rightarrow {b^a} = c\]
Indices formula:
\[{a^{m + n}} = {a^m} \cdot {a^n}\]

Complete step by step solution: Given differential equation is \[\log \left( {\dfrac{{dy}}{{dx}}} \right) = ax + by\]
Applying antilogarithm formula:
\[\dfrac{{dy}}{{dx}} = {e^{ax + by}}\]
Apply indices formula \[{a^{m + n}} = {a^m} \cdot {a^n}\]:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {e^{ax}} \cdot {e^{by}}\]
Separate the variables of the differential equation:
\[ \Rightarrow \dfrac{{dy}}{{{e^{by}}}} = {e^{ax}}dx\]
\[ \Rightarrow {e^{ - by}}dy = {e^{ax}}dx\]
Taking integration on both sides of the differential equation:
\[ \Rightarrow \int {{e^{ - by}}dy} = \int {{e^{ax}}dx} \]
Applying the formula \[\int {{e^{mx}}dx} = \dfrac{{{e^{mx}}}}{m} + c\]:
\[ \Rightarrow \dfrac{{{e^{ - by}}}}{{ - b}} = \dfrac{{{e^{ax}}}}{a} + c\]


Option ‘B’ is correct

Additional Information: In the differential equation, we use e base logarithm when the base is not mentioned in equation. In the given question the base of the logarithm is not mentioned. Thus we assume that the base of the logarithm is e.
To differentiate or integrate a logarithm function, the base of the logarithm must be e. If the base of the logarithm is not e, then we have to change the base.
Note: Students often confused with the indices formula. They used wrong formula \[{a^m} + {a^n} = {a^{m + n}}\]. The correct formula is \[{a^{m + n}} = {a^m} \cdot {a^n}\].