# Can this be solved?

1. Mar 31, 2008

### gamesguru

This is not a homework problem, I just thought of it when I was looking at problems with it being just y and not y^2. Here's the problem. It's entirely possible that it's not solvable, I'm just curious.
$$y'+y^2=x[/itex] 2. Mar 31, 2008 ### HallsofIvy That's non-linear so solving it won't be easy. There is, however, a theorem that says every first order d.e. has an "integrating" factor. You can write that as $dy/dx= x- y^2$ so $dy= (x- y^2)dx$ or $dy+ (y^2- x)dx= 0$. There must exist some function v(x,y) such that $v(x,y)dy+ v(x,y)(y^2-x)dx= 0$ is "exact": that is so that there exist a function f(x,y) so that $df= vdy+ v(y^2-x)dx$. If that is true then we must have $v_x= (v(y^2-x))_y$. But there is no theorem that says it will be easy to find v(x,y)! 3. Apr 2, 2008 ### coomast This equation is of the Riccati type. It can be transformed into a linear one by using the substitution: [tex]y(x)=\frac{1}{u(x)}\cdot \frac{du(x)}{dx}$$

Giving thus as transformed equation:

$$\frac{d^2u}{dx^2}-x\cdot u = 0$$

Which is the one of Airy, with known solution. After the inverse transformation you get the solution of the original equation.