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Question Number 13316 by 433 last updated on 18/May/17

{ ((x′(t)=4x(t)+5y(t))),((y′(t)=4y(t))) :}

$$\begin{cases}{{x}'\left({t}\right)=\mathrm{4}{x}\left({t}\right)+\mathrm{5}{y}\left({t}\right)}\\{{y}'\left({t}\right)=\mathrm{4}{y}\left({t}\right)}\end{cases} \\ $$

Answered by ajfour last updated on 18/May/17

dy=4ydt ∫(dy/y)=4∫dt ln ((y/y_0 ))=4t or y=y_0 e^(4t) ....(i)a ln y=4t+ln y_0 ....(i)b (dy/dx)= ((dy/dt)/(dx/dt)) = ((4y)/(4x+5y)) let y=px ⇒ (dy/dx) = p+x(dp/dx) so, p+x(dp/dx) = ((4y)/(4x+5y))= ((4p)/(4+5p)) x(dp/dx) = ((4p)/(4+5p))−p x(dp/dx) = −((5p^2 )/(4+5p)) ∫ ((5p+4)/(5p^2 ))dp =−∫ (dx/x) ∫ (dp/p) +(4/5)∫ (dp/p^2 ) = −∫ (dx/x) ln p−(4/(5p))=−ln x+C ln y−ln x−((4x)/(5y)) =−ln x +C ((4x)/(5y))=ln y−C and C=ln y_0 −((4x_0 )/(5y_0 )) so, ln ((y/y_0 ))=(4/5)((x/y)−(x_0 /y_0 )) ...(ii) further x=((5y)/4)(ln y−C ) x=(5/4)y_0 e^(4t) (4t+ln y_0 −C ) see (i) =5y_0 te^(4t) +(5/4)y_0 e^(4t) (ln y_0 −C ) x_0 =(5/4)y_0 (ln y_0 −C ) so, x= e^(4t) (x_0 +5y_0 t) .....(iii) y=y_0 e^(4t) .....(i) and ln ((y/y_0 ))=(4/5)((x/y)−(x_0 /y_0 )) ...(ii) might be the required solution.

$${dy}=\mathrm{4}{ydt}\:\:\:\:\: \\ $$$$\int\frac{{dy}}{{y}}=\mathrm{4}\int{dt} \\ $$$$\mathrm{ln}\:\left(\frac{{y}}{{y}_{\mathrm{0}} }\right)=\mathrm{4}{t}\:\:\:\:{or}\:\:\boldsymbol{{y}}=\boldsymbol{{y}}_{\mathrm{0}} \boldsymbol{{e}}^{\mathrm{4}\boldsymbol{{t}}} \:\:....\left({i}\right){a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ln}\:{y}=\mathrm{4}{t}+\mathrm{ln}\:{y}_{\mathrm{0}} \:\:\:\:....\left({i}\right){b} \\ $$$$\frac{{dy}}{{dx}}=\:\frac{{dy}/{dt}}{{dx}/{dt}}\:\:=\:\frac{\mathrm{4}{y}}{\mathrm{4}{x}+\mathrm{5}{y}}\:\:\:\:\:\:\:\:\:\: \\ $$$${let}\:\:{y}={px}\:\:\:\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}\:=\:{p}+{x}\frac{{dp}}{{dx}} \\ $$$${so},\:\:\:\:\:\:{p}+{x}\frac{{dp}}{{dx}}\:=\:\frac{\mathrm{4}{y}}{\mathrm{4}{x}+\mathrm{5}{y}}=\:\frac{\mathrm{4}{p}}{\mathrm{4}+\mathrm{5}{p}} \\ $$$${x}\frac{{dp}}{{dx}}\:=\:\frac{\mathrm{4}{p}}{\mathrm{4}+\mathrm{5}{p}}−{p} \\ $$$${x}\frac{{dp}}{{dx}}\:=\:−\frac{\mathrm{5}{p}^{\mathrm{2}} }{\mathrm{4}+\mathrm{5}{p}} \\ $$$$\int\:\frac{\mathrm{5}{p}+\mathrm{4}}{\mathrm{5}{p}^{\mathrm{2}} }{dp}\:=−\int\:\frac{{dx}}{{x}} \\ $$$$\int\:\frac{{dp}}{{p}}\:+\frac{\mathrm{4}}{\mathrm{5}}\int\:\frac{{dp}}{{p}^{\mathrm{2}} }\:=\:−\int\:\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\:{p}−\frac{\mathrm{4}}{\mathrm{5}{p}}=−\mathrm{ln}\:{x}+{C} \\ $$$$\mathrm{ln}\:{y}−\mathrm{ln}\:{x}−\frac{\mathrm{4}{x}}{\mathrm{5}{y}}\:=−\mathrm{ln}\:{x}\:+{C} \\ $$$$\frac{\mathrm{4}{x}}{\mathrm{5}{y}}=\mathrm{ln}\:{y}−{C}\:\:\:{and}\:\:{C}=\mathrm{ln}\:{y}_{\mathrm{0}} −\frac{\mathrm{4}{x}_{\mathrm{0}} }{\mathrm{5}{y}_{\mathrm{0}} } \\ $$$${so},\:\:\:\:\mathrm{ln}\:\left(\frac{{y}}{{y}_{\mathrm{0}} }\right)=\frac{\mathrm{4}}{\mathrm{5}}\left(\frac{{x}}{{y}}−\frac{{x}_{\mathrm{0}} }{{y}_{\mathrm{0}} }\right)\:\:\:\:...\left({ii}\right) \\ $$$${further} \\ $$$${x}=\frac{\mathrm{5}{y}}{\mathrm{4}}\left(\mathrm{ln}\:{y}−{C}\:\right) \\ $$$${x}=\frac{\mathrm{5}}{\mathrm{4}}{y}_{\mathrm{0}} {e}^{\mathrm{4}{t}} \left(\mathrm{4}{t}+\mathrm{ln}\:{y}_{\mathrm{0}} −{C}\:\right)\:\:\:\:{see}\:\left({i}\right) \\ $$$$\:\:=\mathrm{5}{y}_{\mathrm{0}} {te}^{\mathrm{4}{t}} +\frac{\mathrm{5}}{\mathrm{4}}{y}_{\mathrm{0}} {e}^{\mathrm{4}{t}} \left(\mathrm{ln}\:{y}_{\mathrm{0}} −{C}\:\right) \\ $$$${x}_{\mathrm{0}} =\frac{\mathrm{5}}{\mathrm{4}}{y}_{\mathrm{0}} \left(\mathrm{ln}\:{y}_{\mathrm{0}} −{C}\:\right) \\ $$$${so},\:\:\:\:{x}=\:{e}^{\mathrm{4}{t}} \left({x}_{\mathrm{0}} +\mathrm{5}{y}_{\mathrm{0}} {t}\right)\:\:.....\left({iii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}={y}_{\mathrm{0}} {e}^{\mathrm{4}{t}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....\left({i}\right) \\ $$$${and}\:\:\mathrm{ln}\:\left(\frac{{y}}{{y}_{\mathrm{0}} }\right)=\frac{\mathrm{4}}{\mathrm{5}}\left(\frac{{x}}{{y}}−\frac{{x}_{\mathrm{0}} }{{y}_{\mathrm{0}} }\right)\:\:\:\:...\left({ii}\right) \\ $$$${might}\:{be}\:{the}\:{required}\:{solution}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

Commented by ajfour last updated on 19/May/17

no response..!

$${no}\:{response}..! \\ $$

Commented by 433 last updated on 19/May/17

i would like with eigenvalue and eigenvector

$${i}\:{would}\:{like}\:{with}\:{eigenvalue}\:{and}\:{eigenvector} \\ $$

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