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

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- Thread starter gamesguru
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- #1

- 85

- 2

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

- #2

HallsofIvy

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

- #3

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[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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