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If $\frac{d y}{d x}+2 x \tan (x-y)=1$, then $\sin (x-y)$ is equal to
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The correct answer is:
$A e^{x^2}$
Given differential equation is
$\frac{d y}{d x}+2 x \tan (x-y)=1$
Put $x-y=t$
$\begin{array}{lrl}\Rightarrow & 1-\frac{d y}{d x}=\frac{d t}{d x} \\ \Rightarrow & \frac{d y}{d x}=1-\frac{d t}{d x} \\ \therefore & 1-\frac{d t}{d x}+2 x \tan t=1 \\ \Rightarrow & \frac{d t}{\tan t}=2 x d x \\ \Rightarrow & \cot t d t=2 x d x\end{array}$
On integrating both sides, we get
$\begin{aligned} \log \sin t & =x^2+\log A \\ \Rightarrow \quad \log \frac{\sin (x-y)}{A} & =x^2 \\ \Rightarrow \quad \sin (x-y) & =A e^{x^2}\end{aligned}$
$\frac{d y}{d x}+2 x \tan (x-y)=1$
Put $x-y=t$
$\begin{array}{lrl}\Rightarrow & 1-\frac{d y}{d x}=\frac{d t}{d x} \\ \Rightarrow & \frac{d y}{d x}=1-\frac{d t}{d x} \\ \therefore & 1-\frac{d t}{d x}+2 x \tan t=1 \\ \Rightarrow & \frac{d t}{\tan t}=2 x d x \\ \Rightarrow & \cot t d t=2 x d x\end{array}$
On integrating both sides, we get
$\begin{aligned} \log \sin t & =x^2+\log A \\ \Rightarrow \quad \log \frac{\sin (x-y)}{A} & =x^2 \\ \Rightarrow \quad \sin (x-y) & =A e^{x^2}\end{aligned}$
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