Find the integrating factor for the following differential equation
(cos8x)y′+8(tan8x)y=4sec8x,0<x<π/16.
cos8x y' + 8tan8x y = 4sec8x
as you recall, the IF for
y' + py = q is e^(∫p dx)
Now, if you put your equation into that form, you have
y' + 8(sec8x)(tan8x)y = 4sec^2(8x)
That should look fairly familiar, as far as integrals go.
To find the integrating factor for the given differential equation, we can follow these steps:
Step 1: Write the given differential equation in standard form,
y' + P(x)y = Q(x)
Given differential equation: (cos8x)y' + 8(tan8x)y = 4sec8x
Divide the entire equation by cos(8x) to get:
y' + 8(tan8x)y/cos(8x) = 4sec(8x)/cos(8x)
The equation becomes:
y' + 8(tan8x)y/cos(8x) = 4sec(8x)
Step 2: Identify P(x) and Q(x) from the standard form.
P(x) = 8(tan8x)/cos(8x)
Q(x) = 4sec(8x)
Step 3: The integrating factor (denoted as "IF") can be found using the formula:
IF = e^(∫P(x)dx)
= e^(∫(8tan8x)/cos(8x) dx)
Step 4: Integrate P(x) to find the integrating factor.
Let's integrate P(x):
∫(8tan8x)/cos(8x) dx
To evaluate this integral, we can use the substitution method, where u = sin(8x).
Differentiating u with respect to x, we get du/dx = 8cos(8x).
Rearranging, we have dx = (du / 8cos(8x))
Substituting the values into the integral:
∫(8tan8x)/cos(8x) dx = ∫(8(u/cos(8x)) / (8cos(8x))) du
= ∫(u / cos^2(8x)) du
= ∫(u sec^2(8x)) du
= ∫u sec^2(8x) du
The integral of sec^2(8x) can be evaluated as (1/8)tan(8x) + C, where C represents the constant of integration.
Therefore, integrating P(x) gives us:
∫P(x) dx = (1/8) ∫u sec^2(8x) du
= (1/8) ∫sec^2(8x) du
= (1/8) (1/8)tan(8x) + C
= (1/64)tan(8x) + C
Step 5: The integrating factor (IF) is e^(∫P(x)dx). Therefore, the integrating factor is:
IF = e^((1/64)tan(8x) + C)
Note: The constant of integration (C) will get canceled out later when we multiply the integrating factor with the entire differential equation.
Hence, the integrating factor for the given differential equation is e^((1/64)tan(8x)).