# How to solve the differential equation $\frac{dy}{dx}-\frac{3x^2}{1+x^3}=\frac{\sin^2(x)}{{1+x}}$ and how to find its integrating factor?

How to solve the differential equation $$\frac{dy}{dx}-\frac{3x^2}{1+x^3}=\frac{\sin^2(x)}{{1+x}}$$ and how to find its integrating factor?

The integrating factor is given as $$\frac{1}{1+x^3}$$.Could'nt understand why.Its not even a linear differential equation.

• Why you need an integration factor? the solution is $y=\int\left(\frac{\sin^2(x)}{{1+x}}+\frac{3x^2}{1+x^3}\right)\,dx$ – boaz Jun 19 '16 at 16:39
• You typoed I think. There will be a $y$ beside $3x^2$ and it makes every thing fall into the line.. – Qwerty Jun 19 '16 at 16:43
• Hey what rank did You get . Are you planning to drop or continue? – Archis Welankar Jun 19 '16 at 16:45
• a solution in the known elementary functions doesn't exist – Dr. Sonnhard Graubner Jun 19 '16 at 16:48
• 4341 rank.Yes dropping.Not joining anywhere this year.Maybe i'll get in newer IITS.But no hope of getting good branches in the top 4.So preparing with full determination for 2017.You'll be giving too?Right?Good luck. @ArchisWelankar – user220382 Jun 19 '16 at 17:20