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Above we have the problem in question. It is a non-exact, non-homogeneous question. I already solved the general solution by finding and integrating factor, etc.

The answer to the general homogeneous is sec(x)y + c.

However, I run into trouble finding the particular solutions.

I know for higher order differential equations such as y'' + y'+3y = something.

I can use methods such as undetermined coefficients, variation of parameters, even operational methods! But I have never seen this case.

Please help.

  • 1
    $\begingroup$ Hint: Did you consider Integrating Factor? Divide by $\cos x$ and then look at the form. After, you end up with expressions that can be integrated on the LHS and RHS. $\endgroup$
    – Moo
    Commented Apr 20, 2020 at 20:16
  • $\begingroup$ @Moo, thank you so much. I am in shock. I never expected to be dealing with a separable diff. E. $\endgroup$
    – Josue
    Commented Apr 20, 2020 at 20:23

1 Answer 1


$$\cos (x) y'+ \sin (x) y=2 \cos ^3 x \sin x$$ $$\dfrac {\cos (x) y'+ \sin(x) y}{\cos ^2 x}=2 \cos x \sin x$$ $$\left (\dfrac {y}{\cos (x)} \right )'= \sin (2x)$$ Integrate both sides.

You can also change the variable $$\cos (x) y'+ \sin (x) y=2 \cos ^3 x \sin x$$ $$\cos (x) \dfrac {dy}{d \cos x} \frac {d \cos x}{dx}+ \sin (x) y=2 \cos ^3 x \sin x$$ $$-\sin x \cos (x) \dfrac {dy}{d \cos x} + \sin (x) y=2 \cos ^3 x \sin x$$ $$- \cos (x) \dfrac {dy}{d \cos x} + y=2 \cos ^3 x$$ $$- u \dfrac {dy}{d u} + y=2 u^3$$ Where $u= \cos x$


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