Find the general solution of the following Cauchy-Euler differential equations.

**Link88** · November 23rd 2009, 03:16 PM

2x^3(y''')-3x^2(y'')+5xy'=0

**chisigma** · November 23rd 2009, 11:40 PM

Let's suppose to search as solution of the DE...

$2x^{3}\cdot y^{'''} - 2x^{2}\cdot y^{''} + 5x\cdot y^{'}=0$ (1)

... function of the type $y=x^{\alpha}$ . Is...

$y^{'} = \alpha\cdot x^{\alpha-1}$

$y^{''} = \alpha\cdot (\alpha-1)\cdot x^{\alpha-2}$

$y^{'''} = \alpha\cdot (\alpha-1)\cdot (\alpha-2)\cdot x^{\alpha-3}$ (2)

... and the substitution of (2) in (1) leads us to write...

$2\cdot \alpha\cdot (\alpha-1)\cdot (\alpha-2) - 3\cdot \alpha\cdot (\alpha-1) + 5\cdot \alpha =0$ (3)

If $\alpha_{1}$ , $\alpha_{1}$ and $\alpha_{3}$ are the roots ot the algebraic third order equation (3), then the general solution of (1) will be...

$y= c_{1}\cdot x^{\alpha_{1}} + c_{2}\cdot x^{\alpha_{2}} + c_{3}\cdot x^{\alpha_{3}}$ (4)

Kind regards

$\chi$ $\sigma$

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