# Problem with a first order linear differential equation

I have the following differential equation:

$(y^{2}\cos(x)-3x^2y-8x)dx + (2y\sin(x)-x^3+\ln(y))dy=0$

It's obviously not separable or exact so I'm assuming you solve it like you would a normal linear differential equation but I just can't see how, it's too cumbersome and I can't really figure out where to begin.

Anything helps.

Thank you.

• Maybe try integral by part? $\int y^2cos(x)dx + \int 2ysin(x)dy = y^2sin(x) + C$ for example. – cr001 Sep 15 '17 at 15:02
• It can be written as the sum of an exact differential equation, and a trivial one. $$(y^2\cos(x) - 3x^2y)dx + (2y\sin(x) - x^3)dy = 8xdx - \ln(y)dy$$ – Dhanvi Sreenivasan Sep 15 '17 at 15:17
• "It's obviously not separable or exact": This is an exact differential equation. Please recheck your working. – projectilemotion Sep 15 '17 at 16:20

$$(y^{2}cos(x)-3x^2y-8x)dx + (2ysin(x)-x^3+ln(y))dy=0$$ $$y^2\cos x\,dx-3x^2y\,dx-8x\,dx + 2y\sin x\,dy-x^3\,dy+\ln y\,dy=0$$ $$(y^2\cos x\,dx+ 2y\sin x\,dy)-(3x^2y\,dx+x^3\,dy)-8x\,dx +\ln y\,dy=0$$ $$(y^2d\sin x+ \sin x\,dy^2)-(y\,dx^3+x^3\,dy)-4\,dx^2 +\ln y\,dy=0$$ $$d(\sin x\,y^2)-d(x^3y)-4\,dx^2 +\ln y\,dy=0$$ $$\int\,d(\sin x\,y^2)-\int\,d(x^3y)-4\int\,\,dx^2 +\int\,\ln y\,dy=0$$ $$y^2\sin x\,-x^3y-4x^2 +y\ln y-y=C$$