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


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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

	$d y = 4 y d t$ $\int \frac{d y}{y} = 4 \int d t$ $\ln (\frac{y}{y_{0}}) = 4 t o r y = y_{0} e^{4 t} . . . . (i) a$ $\ln y = 4 t + \ln y_{0} . . . . (i) b$ $\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{4 y}{4 x + 5 y}$ $l e t y = p x \Rightarrow \frac{d y}{d x} = p + x \frac{d p}{d x}$ $s o, p + x \frac{d p}{d x} = \frac{4 y}{4 x + 5 y} = \frac{4 p}{4 + 5 p}$ $x \frac{d p}{d x} = \frac{4 p}{4 + 5 p} - p$ $x \frac{d p}{d x} = - \frac{5 p^{2}}{4 + 5 p}$ $\int \frac{5 p + 4}{5 p^{2}} d p = - \int \frac{d x}{x}$ $\int \frac{d p}{p} + \frac{4}{5} \int \frac{d p}{p^{2}} = - \int \frac{d x}{x}$ $\ln p - \frac{4}{5 p} = - \ln x + C$ $\ln y - \ln x - \frac{4 x}{5 y} = - \ln x + C$ $\frac{4 x}{5 y} = \ln y - C a n d C = \ln y_{0} - \frac{4 x_{0}}{5 y_{0}}$ $s o, \ln (\frac{y}{y_{0}}) = \frac{4}{5} (\frac{x}{y} - \frac{x_{0}}{y_{0}}) . . . (i i)$ $f u r t h e r$ $x = \frac{5 y}{4} (\ln y - C)$ $x = \frac{5}{4} y_{0} e^{4 t} (4 t + \ln y_{0} - C) s e e (i)$ $= 5 y_{0} {t e}^{4 t} + \frac{5}{4} y_{0} e^{4 t} (\ln y_{0} - C)$ $x_{0} = \frac{5}{4} y_{0} (\ln y_{0} - C)$ $s o, x = e^{4 t} (x_{0} + 5 y_{0} t) . . . . . (i i i)$ $y = y_{0} e^{4 t} . . . . . (i)$ $a n d \ln (\frac{y}{y_{0}}) = \frac{4}{5} (\frac{x}{y} - \frac{x_{0}}{y_{0}}) . . . (i i)$ $m i g h t b e t h e r e q u i r e d s o l u t i o n .$
	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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