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.