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.