The general solution of the differential equa | vedclass

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Question

(C) Given the differential equation: $\cos (x+y) \frac{dy}{dx} = 1$. Let $x+y = V$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx}

Vedclass · Accepted Answer

$y = 	an \left(\frac{x+y}{2}ight) + c$

Vedclass · Answer

(C) Given the differential equation: $\cos (x+y) \frac{dy}{dx} = 1$. Let $x+y = V$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dV}{dx}$,which implies $\frac{dy}{dx} = \frac{dV}{dx} - 1$. Substituting these into the original equation: $\cos V \left(\frac{dV}{dx} - 1ight) = 1$. This simplifies to $\cos V \frac{dV}{dx} = 1 + \cos V$. Rearranging the terms for integration: $\int \frac{\cos V}{1 + \cos V} dV = \int dx$. We can rewrite the integrand as $\int \left[ \frac{1 + \cos V}{1 + \cos V} - \frac{1}{1 + \cos V} ight] dV = \int dx$. This becomes $\int dV - \int \frac{1}{2 \cos^2 (V/2)} dV = \int dx$,which is $\int dV - \frac{1}{2} \int \sec^2 (V/2) dV = \int dx$. Integrating both sides: $V - \frac{1}{2} \cdot \frac{	an (V/2)}{1/2} = x + c$. Thus,$V - 	an (V/2) = x + c$. Substituting $V = x+y$ back: $(x+y) - 	an \left(\frac{x+y}{2}ight) = x + c$. Therefore,$y = 	an \left(\frac{x+y}{2}ight) + c$.

The general solution of the differential equation $\cos (x+y) \frac{dy}{dx} = 1$ is

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