# -b.2.2.33 - Homogeneous first order ODEs, direction fields and integral curves

• MHB
• karush
In summary, the equation $x\dfrac{du}{dx}+ u= \dfrac{4u- 3}{2- u}$ has the following solution: $x\dfrac{du}{dx}+ u= \dfrac{4u- 3}{2- u}-u\dfrac{2-u}{2-u}=\dfrac{u^2- 2u-3}{2-u}$.
karush
Gold Member
MHB
$\dfrac{dy}{dx}=\dfrac{4y-3x}{2x-y}$
OK I assume u subst so we can separate
$$\dfrac{dy}{dx}= \dfrac{y/x-3}{2-y/x}$$

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

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

Yes, thanks.

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

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

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

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

this is the book answe... but steps to get there??

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

karush said:
$x\dfrac{du}{dx}+ u= \dfrac{4u- 3}{2- u}$
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
karush said:
how did you get +2u

$\color{red}4u - 2u = +2u$

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

what next?

integrate both sides
but what is the advantage of the factored denominator?

RHS

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karush said:
integrate both sides
but what is the advantage of the factored denominator?
https://www.physicsforums.com/attachments/10197
wow

method of partial fractions

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

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

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

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

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

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

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

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

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

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

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

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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

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

depends which side of 18 you are on

## 1. What is a homogeneous first order ODE?

A homogeneous first order ordinary differential equation (ODE) is a type of differential equation where all the terms can be written as a function of the dependent variable and its derivatives. In other words, there are no constant terms in the equation.

## 2. What is a direction field in the context of ODEs?

A direction field is a graphical representation of a first order ODE that shows the slope or direction of the solution at different points in the solution space. It is created by plotting small line segments with slopes corresponding to the value of the derivative at each point.

## 3. How are integral curves related to first order ODEs?

Integral curves are the solution curves to first order ODEs. They represent the family of all possible solutions to the given ODE and can be found by integrating the equation with respect to the independent variable.

## 4. What is the significance of -b.2.2.33 in the context of ODEs?

The notation -b.2.2.33 is used to represent a specific type of first order ODE, namely a homogeneous first order ODE. The numbers following the b represent the order and type of the equation, which can provide important information about its solution.

## 5. How are direction fields and integral curves used to solve ODEs?

Direction fields and integral curves are used to visualize and understand the behavior of a first order ODE. They can also help in identifying patterns and finding particular solutions to the equation. However, they are not always sufficient for finding exact solutions, and other methods such as separation of variables or variation of parameters may be needed.

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