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Question
Verify that y = A cos x + sin x satisfies the differential equation \[\cos x\frac{dy}{dx} + \left( \sin x \right)y=1.\]
Solution
We have,
\[\cos x\frac{dy}{dx} + \left( \sin x \right)y = 1 . . . . . \left( 1 \right)\]
Now,
y = A cos x + sin x
\[\frac{dy}{dx} = - A \sin x + \cos x\]
\[\text{Putting }\frac{dy}{dx} = - A \sin x + \cos x\text{ and }y = A \cos x + \sin x\text{ in (1), we get}\]
\[\text{LHS }= \left( \cos x \right)\left( - A \sin x + \cos x \right) + \left( \sin x \right) \left( A \cos x + \sin x \right)\]
\[ = - A \sin x \cos x + \cos^2 x + A \cos x \sin x + \sin^2 x\]
\[ = \cos^2 x + \sin^2 x\]
\[ = 1\]
= RHS
Thus, y = A cos x + sin x is the solution of the given differential equation.
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