The differential equation cos (x+y) dy = dx h | vedclass

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Question

(C) Given the differential equation: $\cos (x+y) dy = dx$. Dividing by $dy$,we get $\frac{dx}{dy} = \cos (x+y)$. Let $x+y = u$. Differentiating with r

Vedclass · Accepted Answer

$y = 	an \left(\frac{x+y}{2}ight) + c$,where $c$ is a constant.

Vedclass · Answer

(C) Given the differential equation: $\cos (x+y) dy = dx$. Dividing by $dy$,we get $\frac{dx}{dy} = \cos (x+y)$. Let $x+y = u$. Differentiating with respect to $y$,we get $\frac{dx}{dy} + 1 = \frac{du}{dy}$,which implies $\frac{dx}{dy} = \frac{du}{dy} - 1$. Substituting this into the differential equation: $\frac{du}{dy} - 1 = \cos u$. Rearranging the terms: $\frac{du}{dy} = 1 + \cos u$. Separating the variables: $\frac{du}{1 + \cos u} = dy$. Using the identity $1 + \cos u = 2 \cos^2 \left(\frac{u}{2}ight)$,we get $\frac{du}{2 \cos^2 \left(\frac{u}{2}ight)} = dy$,which simplifies to $\frac{1}{2} \sec^2 \left(\frac{u}{2}ight) du = dy$. Integrating both sides: $\int \frac{1}{2} \sec^2 \left(\frac{u}{2}ight) du = \int dy$. This gives $	an \left(\frac{u}{2}ight) = y + c$. Substituting $u = x+y$ back,we get $	an \left(\frac{x+y}{2}ight) = y + c$,or $y = 	an \left(\frac{x+y}{2}ight) - c$. Since $c$ is an arbitrary constant,we can write the general solution as $y = 	an \left(\frac{x+y}{2}ight) + c$.

The differential equation $\cos (x+y) dy = dx$ has the general solution given by

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