A) \[{{e}^{x}}(\sin x+\cos x)+c=0\]
B) \[{{e}^{y}}(\sin x+\cos x)=c\]
C) \[{{e}^{y}}(\cos x-\sin x)=c\]
D) \[{{e}^{x}}(\sin x-\cos x)=c\]
Correct Answer: B
Solution :
\[\frac{dy}{dx}=-\frac{\cos x-\sin x}{\sin x+\cos x}\] Þ \[dy=-\left( \frac{\cos x-\sin x}{\sin x+\cos x} \right)dx\] On integrating both sides, we get Þ \[y=-\log (\sin x+\cos x)+\log c\] Þ \[y=\log \left( \frac{c}{\sin x+\cos x} \right)\] Þ \[{{e}^{y}}(\sin x+\cos x)=c\].
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