Question Number 126960 by bramlexs22 last updated on 25/Dec/20
$$\:{using}\:{Frobenius}\:{method} \\ $$$${x}^{\mathrm{2}} {y}''\:+\mathrm{6}{xy}'\:+\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right){y}\:=\:\mathrm{0} \\ $$
Commented by liberty last updated on 26/Dec/20
$${put}\:{y}={e}^{{rx}} \:\rightarrow\begin{cases}{{y}'={re}^{{rx}} }\\{{y}''={r}^{\mathrm{2}} {e}^{{rx}} }\end{cases} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{2}} \left({r}^{\mathrm{2}} \:{e}^{{rx}} \:\right)+\:\mathrm{6}{x}\left({r}\:{e}^{{rx}} \right)+\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right){e}^{{rx}} \:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:{e}^{{rx}} \:\left({x}^{\mathrm{2}} {r}^{\mathrm{2}} \:+\mathrm{6}{xr}\:+\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right)\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:{r}^{\mathrm{2}} \:+\:\mathrm{6}{xr}\:+\:\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:{r}\:=\:\frac{−\mathrm{6}{x}\pm\sqrt{\mathrm{36}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right)}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{r}=\:\frac{−\mathrm{6}{x}\:\pm\:\mathrm{2}{x}\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}{x}^{\mathrm{2}} }\:=\:\frac{−\mathrm{3}\pm\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }}{{x}} \\ $$$$\:\:\:\:\:{r}_{\mathrm{1}} =\frac{−\mathrm{3}+\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }}{{x}}\:\wedge\:{r}_{\mathrm{2}} =\:\frac{−\mathrm{3}−\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }}{{x}} \\ $$$${general}\:{solution} \\ $$$$\:\therefore\:{y}\:=\:{C}_{\mathrm{1}} {e}^{−\mathrm{3}+\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }} \:+\:{C}_{\mathrm{2}} {e}^{−\mathrm{3}−\sqrt{\mathrm{3}−\mathrm{4}{x}^{\mathrm{2}} }} \:.\: \\ $$
Commented by bramlexs22 last updated on 26/Dec/20
$${yes}..{thanks}\: \\ $$
Answered by Dwaipayan Shikari last updated on 26/Dec/20
$${y}={e}^{\lambda{x}} \\ $$$$\lambda^{\mathrm{2}} {x}^{\mathrm{2}} {e}^{\lambda{x}} +\mathrm{6}\lambda{e}^{\lambda{x}} +\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}\right){e}^{\lambda{x}} =\mathrm{0} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{6}\lambda{x}+\mathrm{4}{x}^{\mathrm{2}} +\mathrm{6}=\mathrm{0}\Rightarrow\lambda=\frac{−\mathrm{6}{x}\pm\sqrt{\mathrm{36}{x}^{\mathrm{2}} −\mathrm{16}{x}^{\mathrm{4}} −\mathrm{24}{x}^{\mathrm{2}} }}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\lambda=−\frac{\mathrm{3}}{{x}}\pm\sqrt{\frac{\mathrm{3}}{{x}^{\mathrm{2}} }−\mathrm{4}} \\ $$$${y}=\Lambda{e}^{−\frac{\mathrm{3}}{{x}}+\sqrt{\frac{\mathrm{3}}{{x}^{\mathrm{2}} }−\mathrm{4}}} +\Phi{e}^{−\frac{\mathrm{3}}{{x}}−\sqrt{\frac{\mathrm{3}}{{x}^{\mathrm{2}} }−\mathrm{4}}} \\ $$