using-Frobenius-method-x-2-y-6xy-4x-2-6-y-0-

Question Number 126960 by bramlexs22 last updated on 25/Dec/20

u s i n g F r o b e n i u s m e t h o d

x^{2} y^{″} + 6 {x y}^{'} + (4 x^{2} + 6) y = 0

Commented by liberty last updated on 26/Dec/20

p u t y = e^{r x} \to {\begin{cases} y^{'} = {r e}^{r x} \\ y^{″} = r^{2} e^{r x} \end{cases}

\Leftrightarrow x^{2} (r^{2} e^{r x}) + 6 x (r e^{r x}) + (4 x^{2} + 6) e^{r x} = 0

\Leftrightarrow e^{r x} (x^{2} r^{2} + 6 x r + (4 x^{2} + 6)) = 0

\Rightarrow x^{2} r^{2} + 6 x r + (4 x^{2} + 6) = 0

r = \frac{- 6 x \pm \sqrt{36 x^{2} - 4 x^{2} (4 x^{2} + 6)}}{2 x^{2}}

r = \frac{- 6 x \pm 2 x \sqrt{3 - 4 x^{2}}}{2 x^{2}} = \frac{- 3 \pm \sqrt{3 - 4 x^{2}}}{x}

r_{1} = \frac{- 3 + \sqrt{3 - 4 x^{2}}}{x} \land r_{2} = \frac{- 3 - \sqrt{3 - 4 x^{2}}}{x}

g e n e r a l s o l u t i o n

∴ y = C_{1} e^{- 3 + \sqrt{3 - 4 x^{2}}} + C_{2} e^{- 3 - \sqrt{3 - 4 x^{2}}} .

Commented by bramlexs22 last updated on 26/Dec/20

y e s . . t h a n k s

Answered by Dwaipayan Shikari last updated on 26/Dec/20

y = e^{λ x}

λ^{2} x^{2} e^{λ x} + 6 λ e^{λ x} + (4 x^{2} + 6) e^{λ x} = 0

\Rightarrow λ^{2} x^{2} + 6 λ x + 4 x^{2} + 6 = 0 \Rightarrow λ = \frac{- 6 x \pm \sqrt{36 x^{2} - 16 x^{4} - 24 x^{2}}}{2 x^{2}}

λ = - \frac{3}{x} \pm \sqrt{\frac{3}{x^{2}} - 4}

y = Λ e^{- \frac{3}{x} + \sqrt{\frac{3}{x^{2}} - 4}} + Φ e^{- \frac{3}{x} - \sqrt{\frac{3}{x^{2}} - 4}}