-b.2.2.32 First order homogeneous ODE

[karush]

\[ \dfrac{dy}{dx} =\dfrac{x^2+3y^2}{2xy} =\dfrac{x^2}{2xy}+\dfrac{3y^2}{2xy} =\dfrac{x}{2y}+\dfrac{3y}{2x}\]

ok not sure if this is the best first steip,,,, if so then do a $u=\dfrac{x}{y}$ ?

 

[MarkFL]

I would write the ODE as:

\(\displaystyle \frac{dy}{dx}=\frac{1+3\left(\dfrac{y}{x}\right)^2}{2\dfrac{y}{x}}\)

And then use:

\(\displaystyle u=\frac{y}{x}\implies\frac{dy}{dx}=x\frac{du}{dx}+u\)

 

[karush]

$\displaystyle u=\frac{y}{x}\implies\frac{dy}{dx}=x\frac{du}{dx}+u$ so is that $$xu'+x'u=(xu)'$$

 

[karush]

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

 

[MarkFL]

I would write the ODE as:

\(\displaystyle \frac{dy}{dx}=\frac{1+3\left(\dfrac{y}{x}\right)^2}{2\dfrac{y}{x}}\)

And then use:

\(\displaystyle u=\frac{y}{x}\implies\frac{dy}{dx}=x\frac{du}{dx}+u\)

We would then have:

\(\displaystyle x\frac{du}{dx}+u=\frac{1+3u^2}{2u}\)

\(\displaystyle x\frac{du}{dx}=\frac{1+3u^2-2u^2}{2u}\)

\(\displaystyle x\frac{du}{dx}=\frac{1+u^2}{2u}\)

 

[karush]

$x\dfrac{du}{dx}=\dfrac{1+u^2}{2u} $
$x\dfrac{1}{dx}=\dfrac{1+u^2}{2u}\dfrac{1}{du} $
tried to separate varibles but didn't see how it reach the answer of

 

[skeeter]

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

continue ...

 

[MarkFL]

I would next write:

\(\displaystyle \frac{2u}{u^2+1}\,du=\frac{1}{x}\,dx\)

Integrating, we get what?

 

[karush]

$\ln |u^2+1|=\ln|x|$

$u=\dfrac{x}{y}$

 

[MarkFL]

You forgot the constant of integration:

\(\displaystyle \ln|u^2+1|=\ln|cx|\)

And:

\(\displaystyle u=\frac{y}{x}\)

 

[karush]

$\displaystyle \ln|u^2+1|=\ln|cx|$
why did you put the c inside the ln?

so anyway
$\displaystyle \ln\left|\dfrac{u^2+1}{cx}\right|=0$

 

[skeeter]

$\ln(u^2+1) = \ln|x| + C_1$

rewite $C_1=\ln|C_2|$ ...

$\ln\left(\dfrac{y^2}{x^2}+1\right)=\ln|C_2 x|$

note, $\ln{a}=\ln{b} \implies a=b$

$\dfrac{y^2}{x^2} + 1 = Cx$

$y^2+x^2=Cx^3$

 

[karush]

wow... that was a lot of help..

 

[HOI]

$\ln(u^2+1) = \ln|x| + C_1$

rewite $C_1=\ln|C_2|$ ...

$\ln\left(\dfrac{y^2}{x^2}+1\right)=\ln|C_2 x|$

note, $\ln{a}=\ln{b} \implies a=b$

$\dfrac{y^2}{x^2} + 1 = Cx$

$y^2+x^2=Cx^3$

Equivalently
$\ln(u^2+1) = \ln|x| + C_1$
Taking the exponetial of both sides
$e^{ln(u^2+ 1)}= u^2+ 1= e^{ln|x|+ C_1}= e^{ln|x|}e^{C_1}= e^{C_1}|x|$
$u^2+ 1= C_2|x|$ where $C_2= e^{C_1}$.

Now, technically e to any power is positive but we can get the general solution by allowing $C_2$ to be positive or negative and then we no longer need the absolute value: $u^2+ 1= C_2x$.