The general solution of the differential equa | vedclass

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(A) Given differential equation is $\log \left(\frac{dy}{dx}\right) = ax + by$.Taking exponential on both sides,we get $\frac{dy}{dx} = e^{ax + by} =

Vedclass · Accepted Answer

$a e^{-by} + b e^{ax} = c$

Vedclass · Answer

(A) Given differential equation is $\log \left(\frac{dy}{dx}\right) = ax + by$.Taking exponential on both sides,we get $\frac{dy}{dx} = e^{ax + by} = e^{ax} \cdot e^{by}$.Separating the variables,we have $e^{-by} dy = e^{ax} dx$.Integrating both sides,we get $\int e^{-by} dy = \int e^{ax} dx$.This results in $-\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C_1$.Multiplying by $ab$,we get $-a e^{-by} = b e^{ax} + ab C_1$.Rearranging the terms,we get $b e^{ax} + a e^{-by} = c$,where $c = -ab C_1$.Therefore,option $A$ is correct.

The general solution of the differential equation $\log \left(\frac{dy}{dx}\right) = ax + by$ is

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