# Separate equation sin3xdx+2y(cos3x)3dy\sin3x\mathrm{d}x+2y(\cos3x)^3\mathrm{d}y

Find solution by using separable equation sin3xdx+2ycos33xdy=0\sin3x\mathrm{d}x +2y\cos^33x\mathrm{d}y=0

So far I got this

sin3xcos33xdx+2ydy=0\frac{\sin3x}{\cos^33x}\mathrm{d}x + 2y\mathrm{d}y = 0

2ydy=−sin3xcos33xdx 2y\mathrm{d}y = -\frac{\sin3x}{\cos^33x}\mathrm{d}x

Integrate both sides

∫2ydy=−∫sin3xcos33xdx\int2y\mathrm{d}y = -\int\frac{\sin3x}{\cos^33x}\mathrm{d}x

let U=sin3xU=\sin3x and du=3cos3xdx\mathrm{d}u=3\cos3x \mathrm{d}x

y2=∫u(du)3y^2=\int\frac{u}{(\mathrm{d}u)^3}

Now du\mathrm{d}u has become a denominator and cubed, not sure how to proceed, thanks in advance

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Rather try u=cos3xu=\cos 3x. As there is only one dxdx, there can only be one dudu.
– LutzL
2 days ago

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Hint:y2=−∫sin3xcos33xdxy^2=-\int\frac{\sin3x}{\cos^33x}dx
cos3x=u⇒−3sin3xdx=du⇒−sin3xdx=du3\cos 3x=u\Rightarrow-3\sin3xdx=du\Rightarrow-\sin3xdx=\frac{du}{3}
y2=∫du3u3y^2=\int\frac{du}{3u^3}

du/(9u2)=du/(9cos6(3x))du/(9u^2)=du/(9\cos^6(3x)), which is the wrong power. Why did you start with the third power at all? Why not start with the result, i.e., use u=1/cos2(3x)u=1/\cos^2(3x)?
– LutzL
2 days ago

2 days ago

Your first error is in computing −sin3xdx-\sin3x\,dx.
– LutzL
2 days ago

Dear LutzL I think now is correct. Thank you.