The general solution of the differential equa | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the differential equation: $\cos(x+y) dy = dx$. Let $v = x+y$. Then,differentiating with respect to $x$,we get $\frac{dv}{dx} = 1 + \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$. The equation can be rewritten as $\frac{dy}{dx} = \frac{1}{\cos(x+y)} = \sec(x+y)$. Substituting $v$ and $\frac{dy}{dx}$: $\frac{dv}{dx} - 1 = \sec(v)$. $\frac{dv}{dx} = 1 + \sec(v) = 1 + \frac{1}{\cos(v)} = \frac{\cos(v)+1}{\cos(v)}$. Separating variables: $\frac{\cos(v)}{\cos(v)+1} dv = dx$. Using the identity $\cos(v) = 2\cos^2(v/2) - 1$,we have $\frac{2\cos^2(v/2)-1}{2\cos^2(v/2)} dv = dx$. $(1 - \frac{1}{2}\sec^2(v/2)) dv = dx$. Integrating both sides: $\int (1 - \frac{1}{2}\sec^2(v/2)) dv = \int dx$. $v - 	an(v/2) = x + c$. Substituting $v = x+y$: $(x+y) - 	an(\frac{x+y}{2}) = x + c$. $y - 	an(\frac{x+y}{2}) = c$,or $y = 	an(\frac{x+y}{2}) + c$.

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

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