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DE 33 - Homogeneous first order ODEs, direction fields and integral curves

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
Untitled126_20200515070503.png
#33.
OK I assumea u subst so we can separate
$$\dfrac{dy}{dx}= \dfrac{y/x-3}{2-y/x} $$
 
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HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
You have dropped the "4"!

The equation $\frac{dy}{dx}= \frac{4y- 3x}{2x- y}$.

Divide both numerator and denominator by x:
$\frac{dy}{dx}= \frac{4\frac{y}{x}- 2}{2- \frac{y}{x}}$.

Now let $u= \frac{y}{x}$ so that $y= xu$ and $\frac{dy}{dx}= x\frac{du}{dx}+ u$.
The equation becomex $x\frac{du}{dx}+ u= \frac{4u- 2}{2- u}$.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
$\dfrac{dy}{dx} = \dfrac{4 \frac{y}{x} - {\color{red}3}}{2 - \frac{y}{x}}$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Yes, thanks.
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
$x\dfrac{du}{dx}+ u= \dfrac{4u- 3}{2- u}$
so
$x\frac{du}{dx}
=\dfrac{4u- 3}{2- u}-u\dfrac{2-u}{2-u}
=\dfrac{4u-3-2u+u^2}{2-u}
=\dfrac{u^2- 2u-3}{2-u}$
check point
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
$\displaystyle\dfrac{u^2- 2u-3}{2-u}=-\dfrac{(u-2)(u+1)}{u-2} =1¬u$
 
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skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
$\displaystyle\dfrac{u^2- 2u-3}{2-u}=-\dfrac{(u-2)(u+1)}{u-2} =1¬u$
$\displaystyle\dfrac{u^2 {\color{red}+} 2u-3}{2-u}=-\dfrac{(u {\color{red}+3})(u {\color{red}-1})}{u-2}$
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
2020_05_17_12.51.52~2.jpg
this is the book answe..... but steps to get there??
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
$x\dfrac{du}{dx}+ u= \dfrac{4u- 3}{2- u}$
so
$x\frac{du}{dx}
=\dfrac{4u- 3}{2- u}-u\dfrac{2-u}{2-u}
=\dfrac{4u-3-2u+u^2}{2-u}
=\dfrac{u^2- 2u-3}{2-u}$
check point
how did you get +2u
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
$x\dfrac{du}{dx}+ u= \dfrac{4u- 3}{2- u}$
so
$x\frac{du}{dx}
=\dfrac{4u- 3}{2- u}-u\dfrac{2-u}{2-u}
=\dfrac{{\color{red}4u}-3 {\color{red}-2u}+u^2}{2-u}
=\dfrac{u^2- 2u-3}{2-u}$
check point

how did you get +2u
$\color{red}4u - 2u = +2u$
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
OK. I recant... then
$\dfrac{x}{dx} =\dfrac{u^2+2u-3}{2- u} \dfrac{1}{du} $.
or.
$\dfrac{1}{x} dx=\dfrac{2-u}{x^2 +2u-3} du$
 
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skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
$\dfrac{2-u}{(u+3)(u-1)} \, du = \dfrac{dx}{x}$

what next?
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
integrate both sides
but what is the advantage of the factored denominator?
2020_05_17_15.07.34~2.jpg
RHS
 
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skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
integrate both sides
but what is the advantage of the factored denominator?
View attachment 10197
wow
method of partial fractions

$\dfrac{2-u}{(u+3)(u-1)} \, du = \dfrac{dx}{x}$

$\dfrac{2-u}{(u+3)(u-1)} = \dfrac{A}{u+3} + \dfrac{B}{u-1}$

$2-u = A(u-1) + B(u+3)$

$u = 1 \implies B = \dfrac{1}{4}$, $u = -3 \implies A = -\dfrac{5}{4}$

$\dfrac{1}{u-1} - \dfrac{5}{u+3} \, du = \dfrac{4}{x} \, dx$

$\ln\left|\dfrac{u-1}{(u+3)^5}\right| = 4\ln|cx|$

$\left| \dfrac{\frac{y}{x} - 1}{\left(\frac{y}{x}+3\right)^5} \right| = cx^4$

$\left|\frac{y}{x} - 1\right| = cx^4\left|\frac{y}{x} +3\right|^5$

$x \left|\frac{y}{x} - 1\right| = cx^5\left|\frac{y}{x} +3 \right|^5$

$\left| y-x \right| = c\left|y +3x \right|^5$
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
OMG...
OK I see how this works just kinda blind first time through
 
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karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
again thank you so much.
this the last problem on this section. the next one is all. word problems which I hesitate to pursue without more practice
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Isn't "practice" the purpose of the problems?
 

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
depends which side of 18 you are on
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,646
St. Augustine, FL.
Here's a slope field for this problem...move the slider to see some of the solutions...