A particular solution of frac{dy}{dx} = (x + | vedclass

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Question

(A) Given the differential equation $\frac{dy}{dx} = (x + 9y)^2$. Let $v = x + 9y$. Differentiating with respect to $x$,we get $\frac{dv}{dx} = 1 + 9\

Vedclass · Accepted Answer

$3x + 27y = 	an \left[3\left(x + \frac{\pi}{12}ight)ight]$

Vedclass · Answer

(A) Given the differential equation $\frac{dy}{dx} = (x + 9y)^2$. Let $v = x + 9y$. Differentiating with respect to $x$,we get $\frac{dv}{dx} = 1 + 9\frac{dy}{dx}$. Thus,$\frac{dy}{dx} = \frac{1}{9} \left(\frac{dv}{dx} - 1ight)$. Substituting this into the original equation: $\frac{1}{9} \left(\frac{dv}{dx} - 1ight) = v^2$. $\frac{dv}{dx} - 1 = 9v^2 \implies \frac{dv}{dx} = 1 + 9v^2$. Separating the variables: $\frac{dv}{1 + 9v^2} = dx$. Integrating both sides: $\int \frac{dv}{1 + (3v)^2} = \int dx$. $\frac{1}{3} 	an^{-1}(3v) = x + C$. $	an^{-1}(3v) = 3x + 3C$. $3v = 	an(3x + C_1)$,where $C_1 = 3C$. Substituting $v = x + 9y$: $3(x + 9y) = 	an(3x + C_1) \implies 3x + 27y = 	an(3x + C_1)$. Given $x = 0, y = \frac{1}{27}$: $3(0) + 27(\frac{1}{27}) = 	an(3(0) + C_1) \implies 1 = 	an(C_1)$. So,$C_1 = \frac{\pi}{4}$. The particular solution is $3x + 27y = 	an(3x + \frac{\pi}{4})$. Note that $	an(3x + \frac{\pi}{4}) = 	an[3(x + \frac{\pi}{12})]$. Thus,the correct option is $A$.

$A$ particular solution of $\frac{dy}{dx} = (x + 9y)^2$,when $x = 0, y = \frac{1}{27}$ is

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