The general solution of the differential equa | vedclass

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Question

(C) Let $v = 9x + y + 5$. Then differentiating with respect to $x$, we get $\frac{dv}{dx} = 9 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 9$

Vedclass · Accepted Answer

$\frac{1}{3} 	an^{-1} (\frac{9x + y + 5}{3}) = x + c$

Vedclass · Answer

(C) Let $v = 9x + y + 5$. Then differentiating with respect to $x$, we get $\frac{dv}{dx} = 9 + \frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{dv}{dx} - 9$.Substituting into the equation: $\frac{dv}{dx} - 9 = v^2$.$\frac{dv}{dx} = v^2 + 9$.Separating variables: $\int \frac{dv}{v^2 + 3^2} = \int dx$.Using the formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$, we get $\frac{1}{3} 	an^{-1}(\frac{v}{3}) = x + c$.Substituting $v$ back: $\frac{1}{3} 	an^{-1}(\frac{9x + y + 5}{3}) = x + c$.

The general solution of the differential equation $\frac{dy}{dx} = (9x + y + 5)^2$ is:

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