Question Number 219996 by SdC355 last updated on 04/May/25
$$\mathrm{Solve}\:\mathrm{Equation} \\ $$$$\frac{\mathrm{d}{x}\left({t}\right)}{\mathrm{d}{t}}=\mathrm{2}{x}\left({t}\right)+{y}\left({t}\right) \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}{t}}=−\mathrm{3}{y}\left({t}\right) \\ $$$$\begin{pmatrix}{{x}^{\left(\mathrm{1}\right)} \left({t}\right)}\\{{y}^{\left(\mathrm{1}\right)} \left({t}\right)}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix}\begin{pmatrix}{{x}\left({t}\right)}\\{{y}\left({t}\right)}\end{pmatrix} \\ $$$$\mathrm{A}=\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix} \\ $$$$\mathrm{det}\left\{\mathrm{A}−\boldsymbol{\lambda}\mathrm{E}\right\}=\mathrm{0} \\ $$$$\mathrm{det}\left\{\begin{pmatrix}{\mathrm{2}}&{\:\:\:\:\mathrm{1}}\\{\mathrm{0}}&{−\mathrm{3}}\end{pmatrix}−\begin{pmatrix}{\boldsymbol{\lambda}}&{\mathrm{0}}\\{\mathrm{0}}&{\boldsymbol{\lambda}}\end{pmatrix}\right\}=\mathrm{0} \\ $$$$\boldsymbol{\lambda}=\mathrm{2},−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{v}}_{\mathrm{1}} =\begin{pmatrix}{−\mathrm{1}}\\{\:\:\:\:\mathrm{5}}\end{pmatrix} \\ $$$$\boldsymbol{\mathrm{v}}_{\mathrm{2}} =\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}={C}_{\mathrm{1}} {e}^{\lambda{t}} \boldsymbol{\mathrm{v}}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}^{\lambda{t}} \boldsymbol{\mathrm{v}}_{\mathrm{2}} =\begin{pmatrix}{−\mathrm{1}}&{\mathrm{1}}\\{\:\:\:\:\mathrm{5}}&{\mathrm{0}}\end{pmatrix}\begin{pmatrix}{{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} }\\{{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} }\end{pmatrix} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{−{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} +{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} }\\{\:\:\:\:\:\:\:\mathrm{5C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} }\end{pmatrix} \\ $$$$\frac{\mathrm{d}{x}\left({t}\right)}{\mathrm{d}{t}}=−\mathrm{2}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} −\mathrm{3}{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} \neq \\ $$$$\mathrm{2}{x}\left({t}\right)+{y}\left({t}\right)=−\mathrm{2}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} −\mathrm{2}{C}_{\mathrm{2}} {e}^{−\mathrm{3}{t}} +\mathrm{5}{C}_{\mathrm{1}} {e}^{\mathrm{2}{t}} \\ $$$$\frac{\mathrm{d}{y}\left({t}\right)}{\mathrm{d}{t}}=\mathrm{10}{e}^{\mathrm{2}{t}} \neq \\ $$$$−\mathrm{3}{y}\left({t}\right)=−\mathrm{15}{C}_{\mathrm{2}} {e}^{\mathrm{2}{t}} \\ $$$$\:\mathrm{Wrong}…..\:\mathrm{Help} \\ $$
Answered by MrGaster last updated on 04/May/25
