Differential Equation Exercise 16

We want to solve the following first-order differential equation:

$$ y' + y \cos x = \cos x $$

This is a linear differential equation of the form y' + y a(x) = b(x), where \( a(x) = \cos x \) and \( b(x) = \cos x \).

It can be solved using Lagrange’s method, also known as the integrating factor method.

$$ y = e^{- \int a(x) \, dx} \; \left[ \int b(x) \, e^{\int a(x) \, dx} \, dx + c \right] $$

Substituting \( a(x) = \cos x \) and \( b(x) = \cos x \):

$$ y = e^{- \int \cos x \, dx} \; \left[ \int \cos x \; e^{\int \cos x \, dx} \, dx + c \right] $$

The integral of \( \cos x \) is its antiderivative \( \sin x \):

$$ y = e^{-\sin x} \; \left[ \int \cos x \; e^{\sin x} \, dx + c \right] $$

The integral \( \int \cos x \; e^{\sin x} \, dx \) evaluates directly to \( e^{\sin x} \):

$$ y = e^{-\sin x} \; \left[ e^{\sin x} + c \right] $$

Multiplying through by \( e^{-\sin x} \):

$$ y = e^{-\sin x} \cdot e^{\sin x} + c \, e^{-\sin x} $$

$$ y = 1 + c \, e^{-\sin x} $$

Thus, the general solution is:

$$ y = 1 + c \, e^{-\sin x} $$

That completes the solution.