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

using Frobenius method x^2 y′′ +6xy′ +(4x^2 +6)y = 0

$$\:{using}\:{Frobenius}\:{method} \\ $$$${x}^{\mathrm{2}} {y}''\:+\mathrm{6}{xy}'\:+\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right){y}\:=\:\mathrm{0} \\ $$

Commented by liberty last updated on 26/Dec/20

put y=e^(rx) → { ((y′=re^(rx) )),((y′′=r^2 e^(rx) )) :} ⇔ x^2 (r^2 e^(rx) )+ 6x(r e^(rx) )+(4x^2 +6)e^(rx) = 0 ⇔ e^(rx) (x^2 r^2 +6xr +(4x^2 +6)) = 0 ⇒ x^2 r^2 + 6xr + (4x^2 +6) = 0 r = ((−6x±(√(36x^2 −4x^2 (4x^2 +6))))/(2x^2 )) r= ((−6x ± 2x(√(3−4x^2 )))/(2x^2 )) = ((−3±(√(3−4x^2 )))/x) r_1 =((−3+(√(3−4x^2 )))/x) ∧ r_2 = ((−3−(√(3−4x^2 )))/x) general solution ∴ y = C_1 e^(−3+(√(3−4x^2 ))) + C_2 e^(−3−(√(3−4x^2 ))) .

$${put}\:{y}={e}^{{rx}} \:\rightarrow\begin{cases}{{y}'={re}^{{rx}} }\\{{y}''={r}^{\mathrm{2}} {e}^{{rx}} }\end{cases} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{2}} \left({r}^{\mathrm{2}} \:{e}^{{rx}} \:\right)+\:\mathrm{6}{x}\left({r}\:{e}^{{rx}} \right)+\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right){e}^{{rx}} \:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:{e}^{{rx}} \:\left({x}^{\mathrm{2}} {r}^{\mathrm{2}} \:+\mathrm{6}{xr}\:+\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right)\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:{r}^{\mathrm{2}} \:+\:\mathrm{6}{xr}\:+\:\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:{r}\:=\:\frac{−\mathrm{6}{x}\pm\sqrt{\mathrm{36}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right)}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{r}=\:\frac{−\mathrm{6}{x}\:\pm\:\mathrm{2}{x}\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}{x}^{\mathrm{2}} }\:=\:\frac{−\mathrm{3}\pm\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }}{{x}} \\ $$$$\:\:\:\:\:{r}_{\mathrm{1}} =\frac{−\mathrm{3}+\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }}{{x}}\:\wedge\:{r}_{\mathrm{2}} =\:\frac{−\mathrm{3}−\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }}{{x}} \\ $$$${general}\:{solution} \\ $$$$\:\therefore\:{y}\:=\:{C}_{\mathrm{1}} {e}^{−\mathrm{3}+\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }} \:+\:{C}_{\mathrm{2}} {e}^{−\mathrm{3}−\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }} \:.\: \\ $$

Commented by bramlexs22 last updated on 26/Dec/20

yes..thanks

$${yes}..{thanks}\: \\ $$

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

y=e^(λx) λ^2 x^2 e^(λx) +6λe^(λx) +(4x^2 +6)e^(λx) =0 ⇒λ^2 x^2 +6λx+4x^2 +6=0⇒λ=((−6x±(√(36x^2 −16x^4 −24x^2 )))/(2x^2 )) λ=−(3/x)±(√((3/x^2 )−4)) y=Λe^(−(3/x)+(√((3/x^2 )−4))) +Φe^(−(3/x)−(√((3/x^2 )−4)))

$${y}={e}^{\lambda{x}} \\ $$$$\lambda^{\mathrm{2}} {x}^{\mathrm{2}} {e}^{\lambda{x}} +\mathrm{6}\lambda{e}^{\lambda{x}} +\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right){e}^{\lambda{x}} =\mathrm{0} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{6}\lambda{x}+\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}=\mathrm{0}\Rightarrow\lambda=\frac{−\mathrm{6}{x}\pm\sqrt{\mathrm{36}{x}^{\mathrm{2}} −\mathrm{16}{x}^{\mathrm{4}} −\mathrm{24}{x}^{\mathrm{2}} }}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\lambda=−\frac{\mathrm{3}}{{x}}\pm\sqrt{\frac{\mathrm{3}}{{x}^{\mathrm{2}} }−\mathrm{4}} \\ $$$${y}=\Lambda{e}^{−\frac{\mathrm{3}}{{x}}+\sqrt{\frac{\mathrm{3}}{{x}^{\mathrm{2}} }−\mathrm{4}}} +\Phi{e}^{−\frac{\mathrm{3}}{{x}}−\sqrt{\frac{\mathrm{3}}{{x}^{\mathrm{2}} }−\mathrm{4}}} \\ $$

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