Question

Solve DE: (x^2 +1) dy/dx + 3x(y-1) =0, y(0) = 2

Answer 1

I think you can separate and integrate here.

Answer 2

Do some algebra to get all the x's and dx's on one side and all the y's and dy's on the other, then integrate all that stuff.

Answer 3

Separable equation
\[\left(x^2+1\right) \frac{dy}{dx}+3 x (y(x)-1)=0\]
\[\frac{dy}{dx}=-\frac{3 x (y-1)}{x^2+1}\]
\[\frac{1}{y-1}\frac{dy}{dx}\text{ = }-\frac{3 x}{x^2+1}\]
\[\frac{1}{y-1}dy\text{ = }-\frac{3 x}{x^2+1}dx\]
\[\int \frac{1}{y-1}dy\text{ = }-\int\frac{3 x}{x^2+1}dx\]

Answer 4

Thanks guys, I was trying to figure out P(x) the whole time, I guess I was complicating it.

Answer 5

Always look for separation first!