Solve sin(3x)dx+2ycos^3(3x)dy=0 using the separable method

$\sin(3x)dx + 2y\cos^3(3x)dy = 0$

To solve this using the separable method, we can rearrange the terms:

$\sin(3x)dx = -2y\cos^3(3x)dy$

Dividing both sides by $\cos^3(3x)$ , we get:

$\frac{\sin(3x)}{\cos^3(3x)}dx = -2ydy$

To integrate the left-hand side, we can use u-substitution:

Let $u = cos(3x)$ , then $du/dx = -sin(3x)$ and $dx = \frac{du}{-3sin(3x)}$ .

Using trigonometric identities, we can simplify the left-hand side:

$\frac{-sin(3x))}{u^3}.\frac{du}{-3sin(3x))} = 2ydy$

Now, let’s integrate both sides with respect to their respective variables:

$\int \frac{-sin(3x))}{u^3}.\frac{du}{-3sin(3x))} = \int 2ydy$

Substituting in the integral, we get:

$-\frac{1}{6} u^{-2}+C = y^2$

where $C$ is the constant of integration.

By replacing the value of $u=cos(3x)$ , we can simplify the equation as:

$-\frac{1}{6} cos(3x)^{-2}+ C = y^2$

Therefore, the general solution to the differential equation is:

$y=\sqrt{\frac{-1}{6\cos^2(3x}+C}$

where $C$ is the constant of integration.