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Question Number 95447 by Tony Lin last updated on 25/May/20

can I write the solution of ay′′+by′+cy=0 y= { ((c_1 e^(((−b+(√(b^2 −4ac)))/2)x) +c_2 e^(((−b−(√(b^2 −4ac)))/2)x) ,when b^2 −4ac≠0)),((c_1 e^(((−b)/2)x) +c_2 xe^(((−b)/2)x) ,when b^2 −4ac=0)) :} in one sentence not in the form of piecewide-define function

$${can}\:{I}\:{write}\:{the}\:{solution}\:{of} \\ $$$${ay}''+{by}'+{cy}=\mathrm{0} \\ $$$${y}=\begin{cases}{{c}_{\mathrm{1}} {e}^{\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}}{x}} +{c}_{\mathrm{2}} {e}^{\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}}{x}} ,{when}\:{b}^{\mathrm{2}} −\mathrm{4}{ac}\neq\mathrm{0}}\\{{c}_{\mathrm{1}} {e}^{\frac{−{b}}{\mathrm{2}}{x}} +{c}_{\mathrm{2}} {xe}^{\frac{−{b}}{\mathrm{2}}{x}} ,{when}\:{b}^{\mathrm{2}} −\mathrm{4}{ac}=\mathrm{0}}\end{cases} \\ $$$${in}\:{one}\:{sentence} \\ $$$${not}\:{in}\:{the}\:{form}\:{of}\:{piecewide}-{define}\:{function} \\ $$

Answered by mathmax by abdo last updated on 25/May/20

the caracteristic equation is ar^2 +br +c =0 Δ =b^2 −4ac case 1 →b^2 −4ac >0 ⇒r_1 =((−b+(√(b^2 −4ac)))/(2a)) and r_2 =((−b−(√(b^2 −4ax)))/(2a)) ⇒the solution is y(x)=α e^(r_1 x) +β e^(r_2 x) case 2→b^2 −4ac<0 ⇒r_1 =((−b+i(√(4ac−b^2 )))/(2a)) and r_2 =((−b−i(√(4ac−b^2 )))/(2a)) ⇒y(x) =c_1 e^(r_1 x) +c_2 e^(r_2 x) case 3→b^2 −4ac =0 ⇒one roots r =((−b)/(2a)) ⇒ y(x) =(ax+b)e^(rx)

$$\mathrm{the}\:\mathrm{caracteristic}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{ar}^{\mathrm{2}} \:+\mathrm{br}\:+\mathrm{c}\:=\mathrm{0} \\ $$$$\Delta\:=\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}\:\:\:\mathrm{case}\:\mathrm{1}\:\rightarrow\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}\:>\mathrm{0}\:\Rightarrow\mathrm{r}_{\mathrm{1}} =\frac{−\mathrm{b}+\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}}{\mathrm{2a}} \\ $$$$\mathrm{and}\:\mathrm{r}_{\mathrm{2}} =\frac{−\mathrm{b}−\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ax}}}{\mathrm{2a}}\:\Rightarrow\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{y}\left(\mathrm{x}\right)=\alpha\:\mathrm{e}^{\mathrm{r}_{\mathrm{1}} \mathrm{x}} \:+\beta\:\mathrm{e}^{\mathrm{r}_{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{case}\:\mathrm{2}\rightarrow\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}<\mathrm{0}\:\Rightarrow\mathrm{r}_{\mathrm{1}} =\frac{−\mathrm{b}+\mathrm{i}\sqrt{\mathrm{4ac}−\mathrm{b}^{\mathrm{2}} }}{\mathrm{2a}}\:\mathrm{and}\:\mathrm{r}_{\mathrm{2}} =\frac{−\mathrm{b}−\mathrm{i}\sqrt{\mathrm{4ac}−\mathrm{b}^{\mathrm{2}} }}{\mathrm{2a}} \\ $$$$\Rightarrow\mathrm{y}\left(\mathrm{x}\right)\:=\mathrm{c}_{\mathrm{1}} \:\mathrm{e}^{\mathrm{r}_{\mathrm{1}} \mathrm{x}} \:+\mathrm{c}_{\mathrm{2}} \mathrm{e}^{\mathrm{r}_{\mathrm{2}} \mathrm{x}} \:\: \\ $$$$\mathrm{case}\:\mathrm{3}\rightarrow\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}\:=\mathrm{0}\:\Rightarrow\mathrm{one}\:\mathrm{roots}\:\mathrm{r}\:=\frac{−\mathrm{b}}{\mathrm{2a}}\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\left(\mathrm{ax}+\mathrm{b}\right)\mathrm{e}^{\mathrm{rx}} \\ $$

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