Second order ODE reduction to first order solutions

**Jesssa** · March 31st 2012, 11:38 PM

hey,

I have this question I have been trying to solve, I got some help and I think I may have solved it but I just wanted to double check here ,

The question is, Consider the non-linear equation below, use y'=p to reduce it to first order, y(o)=1 and y'(0)=-1

y''+y(y')³=0

$p=\frac{dy}{dx}= \pm\frac{1}{\sqrt{y^2- 2c}}$

so

$\pm dy \sqrt{y^2- 2c}= dx$

Then

$y'(0)=\pm \frac{1}{\sqrt{{{y}^{2}}(0)-2c}}=\pm \frac{1}{\sqrt{1-2c}}=-1$
$\pm \frac{1}{\sqrt{1-2c}}=-1,\,\,\,\pm 1=-\sqrt{1-2c}\,\,squaring\,\,both\,\,sides$
1=1-2c
c=0
$So\,y'(x)=\pm \frac{1}{y}$
$\pm ydy=dx$
$\pm \frac{1}{2}{{y}^{2}}=x+c$
$\pm \frac{1}{2}{{y}^{2}}(0)=\pm \frac{1}{2}=c$
$\therefore y(x)=\sqrt{2x\pm \frac{1}{2}}$

Does this look correct?

Thanks in advance

**princeps** · March 31st 2012, 11:57 PM

Let $y'=v$ , where $v$ is a function with $y$ as argument ,then :

$y''=\frac{dv}{dy} \cdot v=v'_y \cdot v$

Hence :

$y''+(y')^3=0$

$v'_y \cdot v+v^3=0$

$v'_y=-v^2$

$\int \frac{dv}{v^2}=-\int dy$

**Jesssa** · April 1st 2012, 01:25 AM

Isn't that the same result?

you would get

-1/v = -y

so y'= 1/y

only difference is i got +/-, is that the problem in my work having the +/-?

when I did it I had

y'' = dp/dy instead of (dp/dy)p

**princeps** · April 1st 2012, 01:36 AM

I guess that your " $\pm$ " isn't correct...

**Jesssa** · April 1st 2012, 01:47 AM

Ahhh I left out a y!

What a silly mistake

The ODE should be

y''+y(y')3=0

Sorry,

$p'+yp{{'}^{3}}=0$
$\frac{p'}{{{p}^{3}}}=-y$
$\frac{dp}{{{p}^{3}}}=-ydy$
$-\frac{1}{2}\frac{1}{{{p}^{2}}}=-\frac{1}{2}{{y}^{2}}+c$
${{p}^{2}}=\frac{1}{{{y}^{2}}-2c}$
$p=\pm \sqrt{\frac{1}{{{y}^{2}}-2c}}$

Thanks how I got the +/-

**princeps** · April 1st 2012, 02:04 AM

I suggest you to use method that I used..It is a standard method for ODEs of this type .

**Jesssa** · April 1st 2012, 02:25 AM

So using your method the solution would be

$y(x)=\sqrt{2x + \frac{1}{2}}$

Is that right?

Edit:

Hold on that's not right, just working it out quickly now

I get

y³- 3/2 y = 6x - 1/2

Does that look right?

**princeps** · April 1st 2012, 02:50 AM

WA , click "show steps"

**Jesssa** · April 1st 2012, 04:52 AM

...did not know it could do that haha,

so its y³ +5/2 y = 6x + 3/2

thanks!

**princeps** · April 1st 2012, 05:20 AM

You could give a name to this thread "Thanksgiving thread "....

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