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The solution of the differential equation
$\frac{d y}{d x}=\cos (\mathrm{y}-\mathrm{x})+1$ is
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$\frac{d y}{d x}=\cos (\mathrm{y}-\mathrm{x})+1$ is
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Verified Answer
The correct answer is:
$\mathrm{e}^{\mathrm{x}}[\sec (\mathrm{y}-\mathrm{x})-\tan (\mathrm{y}-\mathrm{x})]=\mathrm{c}$
$\frac{d y}{d x}=\cos (y-x)+1$
Let $y-x=t \Rightarrow \frac{d y}{d x}-1=\frac{d t}{d x}$
$\therefore(1) \Rightarrow \frac{d t}{d x}+1=\cos t+1$
$\Rightarrow \frac{d t}{d x}=\cos t$
$\Rightarrow \sec t . d t=1 . d x$
$\Rightarrow \int \sec t \cdot d t=\int 1 \cdot d x$
$\Rightarrow \log |\operatorname{sect}+\tan t|=x+c$
$\Rightarrow-\log |\sec t-\tan t|=x+c$
$\Rightarrow \log |\sec t-\tan t|=-x-c$
$\Rightarrow \log (\sec (y-x)-\tan (y-x)=-x-c$
$\Rightarrow \sec (y-x)-\tan (y-x)=e^{-x} \cdot e^{-c}$
$\Rightarrow e^{x}(\sec (y-x)-\tan (y-x))=c$
Let $y-x=t \Rightarrow \frac{d y}{d x}-1=\frac{d t}{d x}$
$\therefore(1) \Rightarrow \frac{d t}{d x}+1=\cos t+1$
$\Rightarrow \frac{d t}{d x}=\cos t$
$\Rightarrow \sec t . d t=1 . d x$
$\Rightarrow \int \sec t \cdot d t=\int 1 \cdot d x$
$\Rightarrow \log |\operatorname{sect}+\tan t|=x+c$
$\Rightarrow-\log |\sec t-\tan t|=x+c$
$\Rightarrow \log |\sec t-\tan t|=-x-c$
$\Rightarrow \log (\sec (y-x)-\tan (y-x)=-x-c$
$\Rightarrow \sec (y-x)-\tan (y-x)=e^{-x} \cdot e^{-c}$
$\Rightarrow e^{x}(\sec (y-x)-\tan (y-x))=c$
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