The solution of the differential equation log | vedclass

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(B) Given the differential equation: $\log \left(\frac{dy}{dx}\right) = 9x - 6y + 6$. Taking the exponential of both sides,we get: $\frac{dy}{dx} = e^

Vedclass · Accepted Answer

$3e^{6y} = 2e^{9x+6} + e^{6}$

Vedclass · Answer

(B) Given the differential equation: $\log \left(\frac{dy}{dx}\right) = 9x - 6y + 6$. Taking the exponential of both sides,we get: $\frac{dy}{dx} = e^{9x - 6y + 6} = e^{9x+6} \cdot e^{-6y}$. Separating the variables,we have: $e^{6y} dy = e^{9x+6} dx$. Integrating both sides: $\int e^{6y} dy = \int e^{9x+6} dx$. This gives: $\frac{e^{6y}}{6} = \frac{e^{9x+6}}{9} + C$. Given $y = 1$ when $x = 0$: $\frac{e^{6}}{6} = \frac{e^{6}}{9} + C$. Solving for $C$: $C = \frac{e^{6}}{6} - \frac{e^{6}}{9} = \frac{3e^{6} - 2e^{6}}{18} = \frac{e^{6}}{18}$. Substituting $C$ back into the equation: $\frac{e^{6y}}{6} = \frac{e^{9x+6}}{9} + \frac{e^{6}}{18}$. Multiplying by $18$: $3e^{6y} = 2e^{9x+6} + e^{6}$.

The solution of the differential equation $\log \left(\frac{dy}{dx}\right) = 9x - 6y + 6$ is (given that $y = 1$ when $x = 0$):

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