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(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

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

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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