- #1

Benny

- 584

- 0

[tex]\left( {x + y^3 } \right)y' = y[/tex] (treat y as the independent variable).

Answer: Exact equation is [tex]y^{ - 1} \left( {\frac{{dx}}{{dy}}} \right) - xy^{ - 2} = y[/tex] leading to [tex]x = y\left( {k + \frac{{y^2 }}{2}} \right)[/tex].

I'm having trouble indentifying the type of the DE.

[tex]

\left( {x + y^3 } \right)y' = y

[/tex]

[tex]

ydx - \left( {x + y^3 } \right)dy = 0

[/tex]

Writing the equation as I did above makes it kind of look like an exact(or one that can be made exact) DE but I'm still not sure what to do. I'm also being unsure about how to "treat y as the independent variable." All I can gather from that instruction is to find x = x(y). Can someone please help me out?