Can this be solved? Exploring the solvability of [tex]y'+y^2=x[/itex]

[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.
[tex]y'+y^2=x[/itex]

 

[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)!

 

[coomast]

This equation is of the Riccati type. It can be transformed into a linear one by using the substitution:

$$y(x)=\frac{1}{u(x)}\cdot \frac{du(x)}{dx}$$

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.