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Question Number 146771 by ZiYangLee last updated on 15/Jul/21

$$\mathrm{Given}\mathrm{that}{y}^{″}-4y^{\prime }+3y=0,y\left(0\right)=0,y^{\prime }\left(0\right)=2,$$$$\mathrm{find}y\left(\ln 2\right).$$

Answered by mathmax by abdo last updated on 15/Jul/21

$$\to {\mathrm{r}}^2-4\mathrm{r}+3=0\to {\mathrm{\Delta }}^{{}^{\prime }}=4-3=1\Rightarrow {\mathrm{r}}_1=2+1=3\mathrm{and}{\mathrm{r}}_2=2-1=1\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{ae}}^{\mathrm{x}}+{\mathrm{be}}^{3\mathrm{x}}$$$$\mathrm{y}\left(0\right)=0\Rightarrow \mathrm{a}+\mathrm{b}=0\Rightarrow \mathrm{b}=-\mathrm{a}$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)={\mathrm{ae}}^{\mathrm{x}}+{3\mathrm{be}}^{3\mathrm{x}}\mathrm{and}{\mathrm{y}}^{{}^{\prime }}\left(0\right)=2\Rightarrow \mathrm{a}+3\mathrm{b}=2\Rightarrow \mathrm{a}-3\mathrm{a}=2\Rightarrow \mathrm{a}=-1$$$$\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=-{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{3\mathrm{x}}\Rightarrow \mathrm{y}\left(\ln 2\right)=-2+2^3=6$$

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