The solution of the differential equation fra | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = \sin(x+y) 	an(x+y) - 1$Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \fra

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$x + \operatorname{cosec}(x+y) = c$

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(B) Given the differential equation: $\frac{dy}{dx} = \sin(x+y) 	an(x+y) - 1$Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$,which implies $\frac{dy}{dx} = \frac{dz}{dx} - 1$.Substituting this into the original equation:$\frac{dz}{dx} - 1 = \sin z 	an z - 1$$\frac{dz}{dx} = \sin z 	an z = \sin z \cdot \frac{\sin z}{\cos z} = \frac{\sin^2 z}{\cos z}$Rearranging the terms to separate variables:$\int \frac{\cos z}{\sin^2 z} dz = \int dx$Let $\sin z = t$,then $\cos z dz = dt$. The integral becomes:$\int \frac{1}{t^2} dt = x + c$$- \frac{1}{t} = x + c$Since $t = \sin z$,we have $-\operatorname{cosec} z = x + c$,which simplifies to $x + \operatorname{cosec} z = -c$. Replacing $c$ with a new constant $C$,we get:$x + \operatorname{cosec}(x+y) = C$

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

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