Solve the first-order initial value problem using half angle formulas
{y'(x)= 1 + sin^2 x
y (0)= -7
use the identity
cos 2x = 1 - 2sin^2 x
and solving for sin^2 x
sin^2 x = 1/2 - (1/2)cos 2x
then y'(x)= 1 + sin^2 x
becomes
y'(x)= 1 + sin^2 x
becomes
y'(x)= 1 + 1/2 - (1/2)cos 2x
= 3/2 - (1/2)cos 2x
this is now easy to integrate
Don't forget to add the constant.
plug in the point (0,-7) to find that constant.
To solve the first-order initial value problem using half angle formulas, we need to express the given differential equation in terms of half angle formulas, integrate it, and then substitute the initial condition to determine the constant of integration.
Let's begin with the differential equation:
y'(x) = 1 + sin^2(x)
Since we want to use half angle formulas, we can rewrite sin^2(x) as (1 - cos(2x))/2.
y'(x) = 1 + (1 - cos(2x))/2
Now, simplify the equation:
y'(x) = 3/2 - cos(2x)/2
To integrate y'(x), we will treat it as a separable equation:
dy/dx = 3/2 - cos(2x)/2
Separate the variables:
dy = (3 - cos(2x))/2 dx
Now, integrate both sides with respect to their respective variables:
∫dy = (1/2)∫(3 - cos(2x))dx
Integrating the left side gives us just 'y':
y = (1/2)(3x - (1/2)sin(2x)) + C
Let's substitute the initial condition y(0) = -7 to find the value of the constant C:
-7 = (1/2)(0 - (1/2)sin(0)) + C
Simplifying:
-7 = 0 + C
Therefore, the constant of integration C is -7.
Now, we can substitute C back into the equation:
y = (1/2)(3x - (1/2)sin(2x)) - 7
That's the solution to the first-order initial value problem using half angle formulas.