General solution of the differential equation | vedclass

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(C) Given differential equation is $\log \left(\frac{d y}{d x}\right)=a x+b y$.Taking exponential on both sides,we get $\frac{d y}{d x}=e^{a x+b y} =

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$a e^{-b y}+b e^{a x}=c_1$,where $c_1$ is a constant.

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(C) Given differential equation is $\log \left(\frac{d y}{d x}\right)=a x+b y$.Taking exponential on both sides,we get $\frac{d y}{d x}=e^{a x+b y} = e^{a x} \cdot e^{b y}$.Separating the variables,we have $\frac{d y}{e^{b y}} = e^{a x} d x$,which is $e^{-b y} d y = e^{a x} d x$.Integrating both sides,we get $\int e^{-b y} d y = \int e^{a x} d x$.This results in $\frac{e^{-b y}}{-b} = \frac{e^{a x}}{a} + C$.Rearranging the terms,we get $\frac{e^{a x}}{a} + \frac{e^{-b y}}{b} = -C$.Multiplying by $ab$,we get $b e^{a x} + a e^{-b y} = -abC$.Letting $c_1 = -abC$,we obtain $a e^{-b y} + b e^{a x} = c_1$.

General solution of the differential equation $\log \left(\frac{d y}{d x}\right)=a x+b y$ is

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