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Question Number 94969 by bobhans last updated on 22/May/20

$${\mathrm{y}}^{″}-{3\mathrm{y}}^{\prime }+2\mathrm{y}=0;\mathrm{y}=0,{\mathrm{y}}^{\prime }=3\mathrm{for}\mathrm{x}=0$$

Answered by bobhans last updated on 22/May/20

$$\mathrm{homogenous}\mathrm{solution}$$$${\upsilon }^2-3\upsilon +2=0\Rightarrow (\upsilon -2)(\upsilon -1)=0$$$$\{\begin{array}{l}\upsilon =2\\ \upsilon =1\end{array}\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{Ae}}^{2\mathrm{x}}+{\mathrm{Be}}^{\mathrm{x}}$$$$\{\begin{array}{l}0=\mathrm{A}+\mathrm{B}\Rightarrow \mathrm{A}=-\mathrm{B}\\ {\mathrm{y}}^{\prime }={2\mathrm{Ae}}^{2\mathrm{x}}+{\mathrm{Be}}^{\mathrm{x}}\Rightarrow 3=2\mathrm{A}+\mathrm{B}\end{array}$$$$3=-2\mathrm{B}+\mathrm{B}\{\begin{array}{l}\mathrm{B}=-3\\ \mathrm{A}=3\end{array}$$$$\mathrm{y}={3\mathrm{e}}^{2\mathrm{x}}-{3\mathrm{e}}^{\mathrm{x}}$$

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