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Question

(C) We have, $\frac{dy}{dx} = (4x + y + 1)^2$.Let $v = 4x + y + 1$.Then, $\frac{dv}{dx} = 4 + \frac{dy}{dx}$, which implies $\frac{dy}{dx} = \frac{dv}

Vedclass · Accepted Answer

$y = 2 	an(2x + \frac{\pi}{4}) - 4x - 1$

Vedclass · Answer

(C) We have, $\frac{dy}{dx} = (4x + y + 1)^2$.Let $v = 4x + y + 1$.Then, $\frac{dv}{dx} = 4 + \frac{dy}{dx}$, which implies $\frac{dy}{dx} = \frac{dv}{dx} - 4$.Substituting this into the differential equation, we get $\frac{dv}{dx} - 4 = v^2$, or $\frac{dv}{dx} = v^2 + 4$.Separating the variables, we have $\frac{dv}{v^2 + 4} = dx$.Integrating both sides, $\int \frac{dv}{v^2 + 2^2} = \int dx$.This gives $\frac{1}{2} 	an^{-1}(\frac{v}{2}) = x + c$.Substituting $v = 4x + y + 1$, we get $\frac{1}{2} 	an^{-1}(\frac{4x + y + 1}{2}) = x + c$.Given $y(0) = 1$, we substitute $x = 0$ and $y = 1$: $\frac{1}{2} 	an^{-1}(\frac{0 + 1 + 1}{2}) = 0 + c$, so $c = \frac{1}{2} 	an^{-1}(1) = \frac{\pi}{8}$.Thus, $\frac{1}{2} 	an^{-1}(\frac{4x + y + 1}{2}) = x + \frac{\pi}{8}$.Multiplying by $2$, $	an^{-1}(\frac{4x + y + 1}{2}) = 2x + \frac{\pi}{4}$.Taking the tangent of both sides, $\frac{4x + y + 1}{2} = 	an(2x + \frac{\pi}{4})$.Therefore, $y = 2 	an(2x + \frac{\pi}{4}) - 4x - 1$.

The solution of the differential equation $\frac{dy}{dx} = (4x + y + 1)^2$, when $y(0) = 1$ is

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