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

We are given the differential equation:

\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:


where C is the constant of integration.

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