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 $t = x - y$. Then $\frac{dt}{dx} = 1 - \frac{dy}{dx}$, which implies $\frac

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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 $t = x - y$. Then $\frac{dt}{dx} = 1 - \frac{dy}{dx}$, which implies $\frac{dy}{dx} = 1 - \frac{dt}{dx}$. Substituting this into the given equation: $1 - \frac{dt}{dx} + 2x 	an(t) = 1$ $\Rightarrow \frac{dt}{dx} = 2x 	an(t)$ $\Rightarrow \cot(t) dt = 2x dx$. Integrating both sides: $\int \cot(t) dt = \int 2x dx$ $\ln|\sin(t)| = x^2 + C$. Let $C = \ln|A|$, then $\ln|\sin(t)| = x^2 + \ln|A|$. $\ln|\sin(t)| - \ln|A| = x^2$ $\ln|\frac{\sin(t)}{A}| = x^2$ $\sin(t) = A e^{x^2}$. 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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