Verify that y = A cos x + sin x satisfies the differential equation cos $x \frac{d y}{d x} + (\sin x)$ y = 1.

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Solution

We have,
$\cos x \frac{d y}{d x} + (\sin x) y = 1 . . . . . (1)$
Now,
y = A cos x + sin x
^{$\Rightarrow \frac{d y}{d x} = - A \sin x + \cos x Putting \frac{d y}{d x} = - A \sin x + \cos x and y = A \cos x + \sin x in (1), we get$}
$LHS = (\cos x) (- A \sin x + \cos x) + (\sin x) (A \cos x + \sin x) = - A \sin x \cos x + \cos^{2} x + A \cos x \sin x + \sin^{2} x = \cos^{2} x + \sin^{2} x = 1 = RHS$
Thus, y = A cos x + sin x is the solution of the given differential equation.

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