Question Number 13316 by 433 last updated on 18/May/17
$$\begin{cases}{{x}'\left({t}\right)=\mathrm{4}{x}\left({t}\right)+\mathrm{5}{y}\left({t}\right)}\\{{y}'\left({t}\right)=\mathrm{4}{y}\left({t}\right)}\end{cases} \\ $$
Answered by ajfour last updated on 18/May/17
$${dy}=\mathrm{4}{ydt}\:\:\:\:\: \\ $$$$\int\frac{{dy}}{{y}}=\mathrm{4}\int{dt} \\ $$$$\mathrm{ln}\:\left(\frac{{y}}{{y}_{\mathrm{0}} }\right)=\mathrm{4}{t}\:\:\:\:{or}\:\:\boldsymbol{{y}}=\boldsymbol{{y}}_{\mathrm{0}} \boldsymbol{{e}}^{\mathrm{4}\boldsymbol{{t}}} \:\:….\left({i}\right){a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ln}\:{y}=\mathrm{4}{t}+\mathrm{ln}\:{y}_{\mathrm{0}} \:\:\:\:….\left({i}\right){b} \\ $$$$\frac{{dy}}{{dx}}=\:\frac{{dy}/{dt}}{{dx}/{dt}}\:\:=\:\frac{\mathrm{4}{y}}{\mathrm{4}{x}+\mathrm{5}{y}}\:\:\:\:\:\:\:\:\:\: \\ $$$${let}\:\:{y}={px}\:\:\:\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}\:=\:{p}+{x}\frac{{dp}}{{dx}} \\ $$$${so},\:\:\:\:\:\:{p}+{x}\frac{{dp}}{{dx}}\:=\:\frac{\mathrm{4}{y}}{\mathrm{4}{x}+\mathrm{5}{y}}=\:\frac{\mathrm{4}{p}}{\mathrm{4}+\mathrm{5}{p}} \\ $$$${x}\frac{{dp}}{{dx}}\:=\:\frac{\mathrm{4}{p}}{\mathrm{4}+\mathrm{5}{p}}−{p} \\ $$$${x}\frac{{dp}}{{dx}}\:=\:−\frac{\mathrm{5}{p}^{\mathrm{2}} }{\mathrm{4}+\mathrm{5}{p}} \\ $$$$\int\:\frac{\mathrm{5}{p}+\mathrm{4}}{\mathrm{5}{p}^{\mathrm{2}} }{dp}\:=−\int\:\frac{{dx}}{{x}} \\ $$$$\int\:\frac{{dp}}{{p}}\:+\frac{\mathrm{4}}{\mathrm{5}}\int\:\frac{{dp}}{{p}^{\mathrm{2}} }\:=\:−\int\:\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\:{p}−\frac{\mathrm{4}}{\mathrm{5}{p}}=−\mathrm{ln}\:{x}+{C} \\ $$$$\mathrm{ln}\:{y}−\mathrm{ln}\:{x}−\frac{\mathrm{4}{x}}{\mathrm{5}{y}}\:=−\mathrm{ln}\:{x}\:+{C} \\ $$$$\frac{\mathrm{4}{x}}{\mathrm{5}{y}}=\mathrm{ln}\:{y}−{C}\:\:\:{and}\:\:{C}=\mathrm{ln}\:{y}_{\mathrm{0}} −\frac{\mathrm{4}{x}_{\mathrm{0}} }{\mathrm{5}{y}_{\mathrm{0}} } \\ $$$${so},\:\:\:\:\mathrm{ln}\:\left(\frac{{y}}{{y}_{\mathrm{0}} }\right)=\frac{\mathrm{4}}{\mathrm{5}}\left(\frac{{x}}{{y}}−\frac{{x}_{\mathrm{0}} }{{y}_{\mathrm{0}} }\right)\:\:\:\:…\left({ii}\right) \\ $$$${further} \\ $$$${x}=\frac{\mathrm{5}{y}}{\mathrm{4}}\left(\mathrm{ln}\:{y}−{C}\:\right) \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{4}}{y}_{\mathrm{0}} {e}^{\mathrm{4}{t}} \left(\mathrm{4}{t}+\mathrm{ln}\:{y}_{\mathrm{0}} −{C}\:\right)\:\:\:\:{see}\:\left({i}\right) \\ $$$$\:\:=\mathrm{5}{y}_{\mathrm{0}} {te}^{\mathrm{4}{t}} +\frac{\mathrm{5}}{\mathrm{4}}{y}_{\mathrm{0}} {e}^{\mathrm{4}{t}} \left(\mathrm{ln}\:{y}_{\mathrm{0}} −{C}\:\right) \\ $$$${x}_{\mathrm{0}} =\frac{\mathrm{5}}{\mathrm{4}}{y}_{\mathrm{0}} \left(\mathrm{ln}\:{y}_{\mathrm{0}} −{C}\:\right) \\ $$$${so},\:\:\:\:{x}=\:{e}^{\mathrm{4}{t}} \left({x}_{\mathrm{0}} +\mathrm{5}{y}_{\mathrm{0}} {t}\right)\:\:…..\left({iii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}={y}_{\mathrm{0}} {e}^{\mathrm{4}{t}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({i}\right) \\ $$$${and}\:\:\mathrm{ln}\:\left(\frac{{y}}{{y}_{\mathrm{0}} }\right)=\frac{\mathrm{4}}{\mathrm{5}}\left(\frac{{x}}{{y}}−\frac{{x}_{\mathrm{0}} }{{y}_{\mathrm{0}} }\right)\:\:\:\:…\left({ii}\right) \\ $$$${might}\:{be}\:{the}\:{required}\:{solution}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Commented by ajfour last updated on 19/May/17
$${no}\:{response}..! \\ $$
Commented by 433 last updated on 19/May/17
$${i}\:{would}\:{like}\:{with}\:{eigenvalue}\:{and}\:{eigenvector} \\ $$