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Question Number 160659 by ZiYangLee last updated on 04/Dec/21

Given that y′′−4y′+3y=0 and y(0)=0, y′(0)=2, find y(ln 2).

$$\mathrm{Given}\:\mathrm{that}\:{y}''−\mathrm{4}{y}'+\mathrm{3}{y}=\mathrm{0}\:\mathrm{and}\:{y}\left(\mathrm{0}\right)=\mathrm{0}, \\ $$$${y}'\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{find}\:{y}\left(\mathrm{ln}\:\mathrm{2}\right). \\ $$

Answered by mr W last updated on 04/Dec/21

let y=ae^(px) ap^2 e^(px) −4ape^(px) +3ae^(px) =0 a(p^2 −4p+3)e^(px) =0 p^2 −4p+3=0 (p−1)(p−3)=0 ⇒p=1, 3 ⇒y=ae^x +be^(3x) ⇒y′=ae^x +3be^(3x) y(0)=a+b=0 y′(0)=a+3b=2 ⇒a=−1, b=1 ⇒y=−e^x +e^(3x) y(ln 2)=−e^(ln 2) +e^(3ln 2) =−2+8=6

$${let}\:{y}={ae}^{{px}} \\ $$$${ap}^{\mathrm{2}} {e}^{{px}} −\mathrm{4}{ape}^{{px}} +\mathrm{3}{ae}^{{px}} =\mathrm{0} \\ $$$${a}\left({p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{3}\right){e}^{{px}} =\mathrm{0} \\ $$$${p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{3}=\mathrm{0} \\ $$$$\left({p}−\mathrm{1}\right)\left({p}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}=\mathrm{1},\:\mathrm{3} \\ $$$$\Rightarrow{y}={ae}^{{x}} +{be}^{\mathrm{3}{x}} \\ $$$$\Rightarrow{y}'={ae}^{{x}} +\mathrm{3}{be}^{\mathrm{3}{x}} \\ $$$${y}\left(\mathrm{0}\right)={a}+{b}=\mathrm{0} \\ $$$${y}'\left(\mathrm{0}\right)={a}+\mathrm{3}{b}=\mathrm{2} \\ $$$$\Rightarrow{a}=−\mathrm{1},\:{b}=\mathrm{1} \\ $$$$\Rightarrow{y}=−{e}^{{x}} +{e}^{\mathrm{3}{x}} \\ $$$${y}\left(\mathrm{ln}\:\mathrm{2}\right)=−{e}^{\mathrm{ln}\:\mathrm{2}} +{e}^{\mathrm{3ln}\:\mathrm{2}} =−\mathrm{2}+\mathrm{8}=\mathrm{6} \\ $$

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