Looking for some guidance please, I need to find the initial-value problem for dy/dx = 1+2cos^2x / y. y = 1 when x = 0.
any help is much appricated
the hard part is to integrate the cos^2x
we know that cos 2x = 2cos^2x - 1
so replace the 2cos^2x with cos2x + 1
then
dy/dx = 1+2cos^2x
= 1 + cos2x + 1 = 2 + cos2x
then y = 2x + (1/2)sin2x + c
given: when x=0, y = 1
1 = 0 + sin0 + c
c = 1
y = 2x + (1/2)sin2x + 1
To find the initial-value problem for the given differential equation, we need to solve for the particular solution that satisfies the initial condition.
First, let's rewrite the given differential equation in a more standard form. We have:
dy/dx = (1 + 2cos^2(x)) / y
To get the y-term on one side, we can multiply both sides of the equation by y:
y * dy/dx = 1 + 2cos^2(x)
Next, we can rearrange the equation to separate the variables. We move the dx term to the other side of the equation:
y dy = (1 + 2cos^2(x)) dx
Now, we can integrate both sides of the equation. Integrating the left side with respect to y and the right side with respect to x, we have:
∫ y dy = ∫ (1 + 2cos^2(x)) dx
Integrating both sides, we get:
(y^2 / 2) = x + 2x + sin(2x) + C
Where C is the constant of integration.
To find the value of C, we can use the initial condition y = 1 when x = 0:
(1^2 / 2) = 0 + 2(0) + sin(2(0)) + C
1/2 = 0 + 0 + 0 + C
1/2 = C
Therefore, the value of the constant C is 1/2.
Now we can substitute the value of C back into the equation:
(y^2 / 2) = x + 2x + sin(2x) + 1/2
Simplifying the equation, we have:
y^2 = 2x + 4x + 2sin(2x) + 1
Finally, we have found the initial-value problem:
y^2 = 6x + 2sin(2x) + 1
where y = 1 when x = 0.