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- #1

- Thread starter karush
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- #1

- Jan 29, 2012

- 1,151

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 $x\frac{du}{dx}+ u= \frac{4u- 2}{2- u}$.

$\dfrac{dy}{dx} = \dfrac{4 \frac{y}{x} - {\color{red}3}}{2 - \frac{y}{x}}$

- Jan 29, 2012

- 1,151

Yes, thanks.

- Thread starter
- #5

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

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- #6

$\displaystyle\dfrac{u^2- 2u-3}{2-u}=-\dfrac{(u-2)(u+1)}{u-2} =1¬u$

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$\displaystyle\dfrac{u^2 {\color{red}+} 2u-3}{2-u}=-\dfrac{(u {\color{red}+3})(u {\color{red}-1})}{u-2}$$\displaystyle\dfrac{u^2- 2u-3}{2-u}=-\dfrac{(u-2)(u+1)}{u-2} =1¬u$

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- #8

- Thread starter
- #9

how did you get +2u

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

$x\dfrac{du}{dx}+ u= \dfrac{4u- 3}{2- u}$

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

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

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- #11

OK. I recant... then

$\dfrac{x}{dx} =\dfrac{u^2+2u-3}{2- u} \dfrac{1}{du} $.

or.

$\dfrac{1}{x} dx=\dfrac{2-u}{x^2 +2u-3} du$

$\dfrac{x}{dx} =\dfrac{u^2+2u-3}{2- u} \dfrac{1}{du} $.

or.

$\dfrac{1}{x} dx=\dfrac{2-u}{x^2 +2u-3} du$

Last edited:

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

what next?

what next?

- Thread starter
- #13

method of partial fractionsintegrate both sides

but what is the advantage of the factored denominator?

View attachment 10197

wow

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

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

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

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

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

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

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

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

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

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

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- #15

OMG...

OK I see how this works just kinda blind first time through

OK I see how this works just kinda blind first time through

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- #16

this the last problem on this section. the next one is all. word problems which I hesitate to pursue without more practice

- Jan 29, 2012

- 1,151

Isn't "practice" the **purpose** of the problems?

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- #18

depends which side of 18 you are on

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- #19

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