The particular solution of log left(frac{dy}{ | vedclass

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Question

(B) Given the differential equation $\log \left(\frac{dy}{dx}\right) = 3x + 4y$.Taking the exponential on both sides,we get $\frac{dy}{dx} = e^{3x + 4

Vedclass · Accepted Answer

$3e^{-4y} + 4e^{3x} = 7$

Vedclass · Answer

(B) Given the differential equation $\log \left(\frac{dy}{dx}\right) = 3x + 4y$.Taking the exponential on both sides,we get $\frac{dy}{dx} = e^{3x + 4y} = e^{3x} \cdot e^{4y}$.Separating the variables,we have $e^{-4y} \, dy = e^{3x} \, dx$.Integrating both sides,$\int e^{-4y} \, dy = \int e^{3x} \, dx$.This gives $\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C$.Given the initial condition $x = 0$ and $y = 0$,we substitute these values into the equation:$\frac{e^{0}}{-4} = \frac{e^{0}}{3} + C \Rightarrow -\frac{1}{4} = \frac{1}{3} + C$.$C = -\frac{1}{4} - \frac{1}{3} = -\frac{7}{12}$.Substituting $C$ back into the equation: $\frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} - \frac{7}{12}$.Multiplying the entire equation by $-12$: $3e^{-4y} = -4e^{3x} + 7$.Rearranging the terms,we get $3e^{-4y} + 4e^{3x} = 7$.

The particular solution of $\log \left(\frac{dy}{dx}\right) = 3x + 4y$ at $x = y = 0$ is

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