x-t-4x-t-5y-t-y-t-4y-t-

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

{\begin{cases} x^{'} (t) = 4 x (t) + 5 y (t) \\ y^{'} (t) = 4 y (t) \end{cases}

Answered by ajfour last updated on 18/May/17

d y = 4 y d t

\int \frac{d y}{y} = 4 \int d t

\ln (\frac{y}{y_{0}}) = 4 t o r y = y_{0} e^{4 t} \dots . (i) a

\ln y = 4 t + \ln y_{0} \dots . (i) b

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{4 y}{4 x + 5 y}

l e t y = p x \Rightarrow \frac{d y}{d x} = p + x \frac{d p}{d x}

s o, p + x \frac{d p}{d x} = \frac{4 y}{4 x + 5 y} = \frac{4 p}{4 + 5 p}

x \frac{d p}{d x} = \frac{4 p}{4 + 5 p} - p

x \frac{d p}{d x} = - \frac{5 p^{2}}{4 + 5 p}

\int \frac{5 p + 4}{5 p^{2}} d p = - \int \frac{d x}{x}

\int \frac{d p}{p} + \frac{4}{5} \int \frac{d p}{p^{2}} = - \int \frac{d x}{x}

\ln p - \frac{4}{5 p} = - \ln x + C

\ln y - \ln x - \frac{4 x}{5 y} = - \ln x + C

\frac{4 x}{5 y} = \ln y - C a n d C = \ln y_{0} - \frac{4 x_{0}}{5 y_{0}}

s o, \ln (\frac{y}{y_{0}}) = \frac{4}{5} (\frac{x}{y} - \frac{x_{0}}{y_{0}}) \dots (i i)

f u r t h e r

x = \frac{5 y}{4} (\ln y - C)

x = \frac{5}{4} y_{0} e^{4 t} (4 t + \ln y_{0} - C) s e e (i)

= 5 y_{0} {t e}^{4 t} + \frac{5}{4} y_{0} e^{4 t} (\ln y_{0} - C)

x_{0} = \frac{5}{4} y_{0} (\ln y_{0} - C)

s o, x = e^{4 t} (x_{0} + 5 y_{0} t) \dots . . (i i i)

y = y_{0} e^{4 t} \dots . . (i)

a n d \ln (\frac{y}{y_{0}}) = \frac{4}{5} (\frac{x}{y} - \frac{x_{0}}{y_{0}}) \dots (i i)

m i g h t b e t h e r e q u i r e d s o l u t i o n .

Commented by ajfour last updated on 19/May/17

n o r e s p o n s e . .!

Commented by 433 last updated on 19/May/17

i w o u l d l i k e w i t h e i g e n v a l u e a n d e i g e n v e c t o r

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