Can this be solved?

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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]
 

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HallsofIvy
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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 [itex]dy/dx= x- y^2[/itex] so [itex] dy= (x- y^2)dx[/itex] or [itex]dy+ (y^2- x)dx= 0[/itex]. There must exist some function v(x,y) such that [itex]v(x,y)dy+ v(x,y)(y^2-x)dx= 0[/itex] is "exact": that is so that there exist a function f(x,y) so that [itex]df= vdy+ v(y^2-x)dx[/itex]. If that is true then we must have [itex]v_x= (v(y^2-x))_y[/itex].

But there is no theorem that says it will be easy to find v(x,y)!
 
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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}[/tex]

Giving thus as transformed equation:

[tex]\frac{d^2u}{dx^2}-x\cdot u = 0[/tex]

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

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