The general solution of the differential equa | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \cos^2(3x+y)$.Let $3x+y = t$. Then,$3 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac

Vedclass · Accepted Answer

$2\sqrt{3}(x+C)$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = \cos^2(3x+y)$.Let $3x+y = t$. Then,$3 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 3$.Substituting this into the equation: $\frac{dt}{dx} - 3 = \cos^2 t$,which gives $\frac{dt}{dx} = \cos^2 t + 3$.Separating variables: $\frac{dt}{\cos^2 t + 3} = dx$.Integrating both sides: $\int \frac{dt}{\cos^2 t + 3} = \int dx$.Multiply numerator and denominator by $\sec^2 t$: $\int \frac{\sec^2 t dt}{1 + 3\sec^2 t} = \int dx$.Using $\sec^2 t = 1 + 	an^2 t$: $\int \frac{\sec^2 t dt}{1 + 3(1 + 	an^2 t)} = \int \frac{\sec^2 t dt}{4 + 3	an^2 t} = \int dx$.Let $	an t = m$,then $\sec^2 t dt = dm$.The integral becomes $\int \frac{dm}{4 + 3m^2} = \frac{1}{3} \int \frac{dm}{\frac{4}{3} + m^2} = \frac{1}{3} \cdot \frac{1}{\frac{2}{\sqrt{3}}} 	an^{-1}\left(\frac{m}{2/\sqrt{3}}ight) = x + C$.Simplifying: $\frac{1}{2\sqrt{3}} 	an^{-1}\left(\frac{\sqrt{3}m}{2}ight) = x + C$.Multiplying by $2\sqrt{3}$: $	an^{-1}\left(\frac{\sqrt{3}}{2} 	an tight) = 2\sqrt{3}(x+C)$.Since $t = 3x+y$,we have $	an^{-1}\left(\frac{\sqrt{3}}{2} 	an(3x+y)ight) = 2\sqrt{3}(x+C)$.Thus,$f(x) = 2\sqrt{3}(x+C)$.

The general solution of the differential equation $\frac{dy}{dx} = \cos^2(3x+y)$ is $\tan^{-1}\left(\frac{\sqrt{3}}{2} \tan(3x+y)\right) = f(x)$. Then,$f(x) =$

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