The general solution of the differential equa | vedclass

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(B) 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} -

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) 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 this into the given differential equation: $\frac{dv}{dx} - 1 = \cos v$. Rearranging the terms,we get $\frac{dv}{dx} = 1 + \cos v$. Using the trigonometric identity $1 + \cos v = 2 \cos^2 \left(\frac{v}{2}ight)$,we have $\frac{dv}{dx} = 2 \cos^2 \left(\frac{v}{2}ight)$. Separating the variables: $\frac{dv}{2 \cos^2 \left(\frac{v}{2}ight)} = dx$,which simplifies to $\frac{1}{2} \sec^2 \left(\frac{v}{2}ight) dv = dx$. Integrating both sides: $\int \frac{1}{2} \sec^2 \left(\frac{v}{2}ight) dv = \int dx$. This gives $	an \left(\frac{v}{2}ight) = x + c$. Substituting $v = x+y$ back,we get $	an \left(\frac{x+y}{2}ight) = x + c$.

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

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