Solve

**dhiab** · December 7th 2009, 03:10 AM

Solve this diferntiel equation :
${y}'=y+y^{3}$

$x{{y}'=y+\sqrt{x^{2}+y^{2}}}$

**mr fantastic** · December 7th 2009, 03:45 AM

Originally Posted by dhiab

Solve this diferntiel equation :
${y}'=y+y^{3}$

$x{{y}'=y+\sqrt{x^{2}+y^{2}}}$

1) $\frac{dy}{dx} = y + y^3 \Rightarrow \int \frac{dy}{y + y^3} = \int \, dx$ . Do the left hand integral by using partial fractions.

2) The DE re-arranges into $\frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \left( \frac{y}{x}\right)^2}$ and the standard technique is to make the substitution $y = xv$ .

If you need more help, please post all your work and say where you are stuck.

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