If the solution of frac{dy}{dx} = (3x + y + 4 | vedclass

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Question

(C) Given the differential equation $\frac{dy}{dx} = (3x + y + 4)^2$. Let $3x + y + 4 = t$. Differentiating with respect to $x$,we get $3 + \frac{dy}{

Vedclass · Accepted Answer

$3\sqrt{3}$

Vedclass · Answer

(C) Given the differential equation $\frac{dy}{dx} = (3x + y + 4)^2$. Let $3x + y + 4 = t$. Differentiating with respect to $x$,we get $3 + \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = \frac{dt}{dx} - 3$. Substituting this into the differential equation: $\frac{dt}{dx} - 3 = t^2$. $\frac{dt}{dx} = t^2 + 3$. Separating the variables: $\frac{dt}{t^2 + 3} = dx$. Integrating both sides: $\int \frac{dt}{t^2 + (\sqrt{3})^2} = \int dx$. Using the formula $\int \frac{dt}{t^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{t}{a}) + C$,we get $\frac{1}{\sqrt{3}} 	an^{-1}(\frac{t}{\sqrt{3}}) = x + k$. Substituting $t = 3x + y + 4$ back: $\frac{1}{\sqrt{3}} 	an^{-1}(\frac{3x + y + 4}{\sqrt{3}}) - x = k$. Comparing this with the given form $\frac{1}{\sqrt{3}} 	an^{-1}(f(x, y)) - x = k$,we identify $f(x, y) = \frac{3x + y + 4}{\sqrt{3}}$. Now,calculate $f(1, 2) = \frac{3(1) + 2 + 4}{\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3}$.

If the solution of $\frac{dy}{dx} = (3x + y + 4)^2$ is $\frac{1}{\sqrt{3}} \tan^{-1}(f(x, y)) - x = k$,then $f(1, 2) = $

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