Solving the Differential Equation (y cos x y cos x)

Let's solve the differential equation (y cos x y cos x) step by step. We will explore both the homogeneous and non-homogeneous cases.

Homogeneous Equation: (y' cos x y 0)

The homogeneous equation is simpler and can be solved directly. We can write it as:

[y' cos x y 0]

Rearranging, we get:

[y' y sec x 0]

This is a first-order linear differential equation. We can solve it using an integrating factor. The integrating factor (I.F) is given by:

[I.F e^{int sec x , dx} e^{ln (sec x tan x)} sec x tan x]

Thus, multiplying the entire equation by the integrating factor gives:

[ (y(sec x tan x))' 0 ]

Integrating both sides, we obtain:

[ y(sec x tan x) C ]

Dividing through by the integrating factor, we find:

[ y frac{C}{sec x tan x} ]

Non-Homogeneous Equation: (y' cos x y cos x)

Now, let's consider the non-homogeneous equation:

[y' cos x y cos x]

We can rewrite it in the standard form

[y' frac{cos x}{cos x} y 1 Rightarrow y' y sec x 1]

Here, we again have a first-order linear differential equation. The integrating factor is:

[I.F e^{int sec x , dx} e^{ln (sec x tan x)} sec x tan x]

Multiplying the entire equation by the integrating factor, we get:

[ (sec x tan x) y' y (sec x tan x) sec x sec x tan x ]

Which simplifies to:

[ frac{d}{dx} [y (sec x tan x)] sec x tan x ]

Integrating both sides, we obtain:

[ y (sec x tan x) int (sec x tan x) , dx ]

The integral of

[ (sec x tan x) ln |sec x tan x| C ]

Therefore, we get:

[ y (sec x tan x) ln |sec x tan x| C ]

Dividing by the integrating factor, we obtain the general solution:

[ y frac{ln |sec x tan x| C}{sec x tan x} ]

Verification of Solutions

We can verify that both solutions satisfy the original equation:

1. For the homogeneous equation:

[ y frac{C}{sec x tan x} ]

Substituting this into the original equation, the terms cancel out, confirming the solution.

2. For the non-homogeneous equation:

[ y frac{ln |sec x tan x| C}{sec x tan x} ]

Substituting and simplifying, we see that the left-hand side indeed equals the right-hand side, confirming this as a solution to the original differential equation.

As a final note, if we consider specific solutions like (y 1), it can be a particular solution as well, which can be verified by substitution.