Question

dy/dx = (1+2cos^2(x))/y, (y > 0), y = 1 when x = 0

Answer 1

To solve the differential equation

\[
\frac{dy}{dx} = \frac{1 + 2\cos^2(x)}{y},
\]

we can start by rewriting it in a separable form. Rearranging gives us:

\[
y \, dy = (1 + 2\cos^2(x)) \, dx.
\]

Now, we can integrate both sides. The left side becomes:

\[
\int y \, dy = \frac{1}{2} y^2.
\]

For the right side, we need to integrate \(1 + 2\cos^2(x)\):

\[
\int (1 + 2\cos^2(x)) \, dx.
\]

We can use the identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\) to rewrite \(2\cos^2(x)\):

\[
2\cos^2(x) = 1 + \cos(2x).
\]

Thus,

\[
1 + 2\cos^2(x) = 2 + \cos(2x).
\]

Now integrating this term by term:

\[
\int (2 + \cos(2x)) \, dx = 2x + \frac{1}{2} \sin(2x) + C,
\]

where \(C\) is a constant of integration.

Putting it all together, we have:

\[
\frac{1}{2} y^2 = 2x + \frac{1}{2} \sin(2x) + C.
\]

To eliminate the \( \frac{1}{2} \) factor, we multiply the whole equation by 2:

\[
y^2 = 4x + \sin(2x) + 2C.
\]

Let \(2C\) be a new constant, which we can call \(C_1\):

\[
y^2 = 4x + \sin(2x) + C_1.
\]

Next, we apply the initial condition \(y(0) = 1\). Plugging \(x = 0\) and \(y = 1\) into the equation gives:

\[
1^2 = 4(0) + \sin(0) + C_1 \implies 1 = 0 + 0 + C_1 \implies C_1 = 1.
\]

Now substituting \(C_1\) back into the equation gives:

\[
y^2 = 4x + \sin(2x) + 1.
\]

Taking the square root (and since \(y > 0\)) results in:

\[
y = \sqrt{4x + \sin(2x) + 1}.
\]

Thus, the solution to the differential equation with the given initial condition is:

\[
\boxed{y = \sqrt{4x + \sin(2x) + 1}}.
\]