# <help> Differential Equations... MAPS (urgent)

• Aug 10th 2007, 11:55 PM
jungohyj
<help> Differential Equations... MAPS (urgent)
There are some differential equations that require other techniques in order to solve them. One type, called homogenous, requires a substitution in order to solve.
The substitution used can be of the form y = v x , or y = v / x,
where v is a function of x.

By using the substituion y = v x, and then the substitution
$\displaystyle u^{2} = 1 + v^{2}$, find the general solution of the differential equation $\displaystyle y\frac{dy}{dx} = \left[ -x + \sqrt{x^{2} + y^{2}}\right]$
• Aug 11th 2007, 02:35 AM
topsquark
Quote:

Originally Posted by jungohyj
There are some differential equations that require other techniques in order to solve them. One type, called homogenous, requires a substitution in order to solve.
The substitution used can be of the form y = v x , or y = v / x,
where v is a function of x.

By using the substituion y = v x, and then the substitution
$\displaystyle u^{2} = 1 + v^{2}$, find the general solution of the differential equation $\displaystyle y\frac{dy}{dx} = \left[ -x + \sqrt{x^{2} + y^{2}}\right]$

First, $\displaystyle y = vx \implies \frac{dy}{dx} = \frac{dv}{dx}x + v$, so
$\displaystyle y\frac{dy}{dx} = \left[ -x + \sqrt{x^{2} + y^{2}}\right]$
becomes
$\displaystyle (vx) \left ( \frac{dv}{dx}x + v \right ) = \left [ -x + \sqrt{x^2 + (vx)^2} \right ]$

$\displaystyle x^2 v \frac{dv}{dx} + xv^2 = \left [ -x + \sqrt{x^2 + x^2v^2} \right ]$

$\displaystyle x^2 v \frac{dv}{dx} + xv^2 = \left [ -x + x \sqrt{1 + v^2} \right ]$

$\displaystyle x \left ( xv \frac{dv}{dx} + v^2 \right ) = x \left [ -1 + \sqrt{1 + v^2} \right ]$

$\displaystyle xv \frac{dv}{dx} + v^2 = \left [ -1 + \sqrt{1 + v^2} \right ]$

Now let $\displaystyle u^2 = 1 + v^2 \implies 2u \frac{du}{dx} = 2v \frac{dv}{dx}$ or $\displaystyle u \frac{du}{dx} = v \frac{dv}{dx}$

Thus
$\displaystyle xu \frac{du}{dx} + (u^2 - 1)= \left [ -1 + \sqrt{u^2} \right ]$

$\displaystyle xu \frac{du}{dx} = -(u^2 - 1) + (|u| - 1)$

$\displaystyle xu \frac{du}{dx} = -u^2 + |u|$

$\displaystyle -\frac{u}{u^2 - |u|}du = \frac{dx}{x}$

So this is either:
$\displaystyle -\frac{1}{u - 1}du = \frac{dx}{x}$ if u > 0
or
$\displaystyle -\frac{1}{u + 1}du = \frac{dx}{x}$ if u < 0

I'm sure you can take the solution from here.

Note: Earlier we divided both sides of the equation by x, so we need to exclude x = 0 from our solution.

-Dan
• Aug 11th 2007, 04:15 AM
Soroban
Hello, jungohyj!

I used a slightly different approach and got a slightly different result . . .

Quote:

By using the substituion $\displaystyle y \,= \,vx$, and then the substitution $\displaystyle u^{2} = 1 + v^{2}$,
find the general solution of: .$\displaystyle y\frac{dy}{dx} = \left[ -x + \sqrt{x^{2} + y^{2}}\right]$

Divide by $\displaystyle x\!:\;\;\frac{y}{x}\!\cdot\!\frac{dy}{dx} \;=\;-\frac{x}{x} + \frac{\sqrt{x^2+y^2}}{x}\;=\;\sqrt{1 + \left(\frac{y}{x}\right)^2} - 1$ . . Note: $\displaystyle x \neq 0$

Let: $\displaystyle v = \frac{y}{x}\quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\!\cdot\!\frac{dv}{dx}$

Substitute: .$\displaystyle v\left(v + x\!\cdot\!\frac{dv}{dx}\right) \;=\;\sqrt{1 + v^2} - 1\quad\Rightarrow\quad v^2 + vx\!\cdot\!\frac{dv}{dx} \;=\;\sqrt{1+v^2} - 1$

and we have: .$\displaystyle vx\!\cdot\!\frac{dv}{dx}\;=\;\sqrt{1+v^2} - (1 + v^2)\quad\Rightarrow\quad vx\!\cdot\!\frac{dv}{dx} \;=\;\sqrt{1+v^2}\left(1 - \sqrt{1+v^2}\right)$

. . $\displaystyle \Rightarrow\quad x(v\,dv) \;=\;\sqrt{1+v^2}\left(1 - \sqrt{1+v^2}\right)\,dx$

Let $\displaystyle u \:=\:\sqrt{1+v^2}\quad\Rightarrow\quad v^2 \:=\:u^2-1\quad\Rightarrow\quad v\,dv \:=\:u\,du$

Substitute: .$\displaystyle x(u\,du) \;=\;u(1 - u)\,dx$

. . and we have: .$\displaystyle x\,du \;=\;(1 - u)\,dx\quad\Rightarrow\quad \frac{du}{1-u} \:=\:\frac{dx}{x}$

• Aug 11th 2007, 04:44 AM
topsquark
Quote:

Originally Posted by Soroban
Let $\displaystyle u \:=\:\sqrt{1+v^2}\quad\Rightarrow\quad v^2 \:=\:u^2-1\quad\Rightarrow\quad v\,dv \:=\:u\,du$

But
$\displaystyle u^2 = 1 + v^2 \implies |u| = \sqrt{1 + v^2}$ in general, because $\displaystyle \sqrt{u^2} = |u|$.

I don't see how you are getting around the absolute value. :confused:

-Dan