Previous Year NEET Question- Differential Equations

NEET 2019: Solve the differential equation $\dfrac{dy}{dx} + y = \cos x$.

Solution:

The given differential equation is of the form $\dfrac{dy}{dx} + Py = Q$, where $P = 1$ and $Q = \cos x$. The integrating factor is $\mu(x) = e^{\int P dx} = e^x$.

Multiplying both sides of the differential equation by $\mu(x)$, we get

$$ e^x \dfrac{dy}{dx} + e^x y = e^x \cos x $$

or, $\dfrac{d}{dx}(e^x y) = e^x \cos x$

Integrating both sides, we get

$$ e^x y = \int e^x \cos x dx + C $$

$$ e^x y = e^x \sin x + C $$

Dividing both sides by $e^x$, we get

$$ y = \sin x + C e^{-x}