If frac{dy}{dx} + 2x tan(x-y) = 1,then sin(x- | vedclass

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(C) Given differential equation is $\frac{dy}{dx} + 2x 	an(x-y) = 1$.Let $x-y = t$.Then $1 - \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{

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$A e^{x^2}$

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(C) Given differential equation is $\frac{dy}{dx} + 2x 	an(x-y) = 1$.Let $x-y = t$.Then $1 - \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dt}{dx}$.Substituting this into the original equation:$1 - \frac{dt}{dx} + 2x 	an(t) = 1$.This simplifies to $\frac{dt}{dx} = 2x 	an(t)$,or $\frac{dt}{	an(t)} = 2x dx$.Integrating both sides: $\int \cot(t) dt = \int 2x dx$.$\ln|\sin(t)| = x^2 + C$.Let $C = \ln|A|$,so $\ln|\sin(t)| = x^2 + \ln|A|$.$\ln|\frac{\sin(t)}{A}| = x^2$.$\sin(t) = A e^{x^2}$.Substituting $t = x-y$ back,we get $\sin(x-y) = A e^{x^2}$.

If $\frac{dy}{dx} + 2x \tan(x-y) = 1$,then $\sin(x-y)$ is equal to

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