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(B) Given differential equation is $(\sin x + \cos x)dy + (\cos x - \sin x)dx = 0$.Rearranging the terms,we get $(\sin x + \cos x)dy = -(\cos x - \sin

Vedclass · Accepted Answer

$e^y(\sin x + \cos x) = c$

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(B) Given differential equation is $(\sin x + \cos x)dy + (\cos x - \sin x)dx = 0$.Rearranging the terms,we get $(\sin x + \cos x)dy = -(\cos x - \sin x)dx$.Thus,$dy = -\frac{\cos x - \sin x}{\sin x + \cos x} dx$.Integrating both sides,$\int dy = -\int \frac{\cos x - \sin x}{\sin x + \cos x} dx$.Let $u = \sin x + \cos x$,then $du = (\cos x - \sin x)dx$.So,$\int dy = -\int \frac{1}{u} du$.$y = -\ln|\sin x + \cos x| + C$.$y = \ln|\frac{c}{\sin x + \cos x}|$ (where $C = \ln c$).Taking exponential on both sides,$e^y = \frac{c}{\sin x + \cos x}$.Therefore,$e^y(\sin x + \cos x) = c$.

The solution of the differential equation $(\sin x + \cos x)dy + (\cos x - \sin x)dx = 0$ is

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