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Question

(A) Given differential equation is $\cos(x+y) dy = dx$,which can be written as $\frac{dy}{dx} = \sec(x+y)$.Let $x+y = t$. Then $1 + \frac{dy}{dx} = \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given differential equation is $\cos(x+y) dy = dx$,which can be written as $\frac{dy}{dx} = \sec(x+y)$.Let $x+y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.Substituting this into the equation: $\frac{dt}{dx} - 1 = \sec(t) \Rightarrow \frac{dt}{dx} = 1 + \sec(t) = \frac{1+\cos(t)}{\cos(t)}$.Separating the variables: $\int \frac{\cos(t)}{1+\cos(t)} dt = \int dx$.Using the identity $1+\cos(t) = 2\cos^2(t/2)$,we get $\int \frac{\cos(t)}{2\cos^2(t/2)} dt = \int dx$.Since $\cos(t) = 2\cos^2(t/2) - 1$,the integral becomes $\int \frac{2\cos^2(t/2)-1}{2\cos^2(t/2)} dt = \int dx$.This simplifies to $\int (1 - \frac{1}{2}\sec^2(t/2)) dt = \int dx$.Integrating both sides: $t - 	an(t/2) = x + C$.Substituting $t = x+y$: $(x+y) - 	an((x+y)/2) = x + C \Rightarrow y - 	an((x+y)/2) = C$.Given $y(0) = 0$,we have $0 - 	an(0/2) = C \Rightarrow C = 0$.Thus,the solution is $y = 	an \left(\frac{x+y}{2}ight)$.

The solution of the differential equation $\cos(x+y) dy = dx$ given that $y(0) = 0$ is

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