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

c a n I w r i t e t h e s o l u t i o n o f

{a y}^{″} + {b y}^{'} + c y = 0

y = {\begin{cases} c_{1} e^{\frac{- b + \sqrt{b^{2} - 4 a c}}{2} x} + c_{2} e^{\frac{- b - \sqrt{b^{2} - 4 a c}}{2} x}, w h e n b^{2} - 4 a c \neq 0 \\ c_{1} e^{\frac{- b}{2} x} + c_{2} {x e}^{\frac{- b}{2} x}, w h e n b^{2} - 4 a c = 0 \end{cases}

i n o n e s e n t e n c e

n o t i n t h e f o r m o f p i e c e w i d e - d e f i n e f u n c t i o n

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

the caracteristic equation is {ar}^{2} + br + c = 0

Δ = b^{2} - 4 ac case 1 \to b^{2} - 4 ac > 0 \Rightarrow r_{1} = \frac{- b + \sqrt{b^{2} - 4 ac}}{2 a}

and r_{2} = \frac{- b - \sqrt{b^{2} - 4 ax}}{2 a} \Rightarrow the solution is y (x) = α e^{r_{1} x} + β e^{r_{2} x}

case 2 \to b^{2} - 4 ac < 0 \Rightarrow r_{1} = \frac{- b + i \sqrt{4 ac - b^{2}}}{2 a} and r_{2} = \frac{- b - i \sqrt{4 ac - b^{2}}}{2 a}

\Rightarrow y (x) = c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}

case 3 \to b^{2} - 4 ac = 0 \Rightarrow one roots r = \frac{- b}{2 a} \Rightarrow

y (x) = (ax + b) e^{rx}