separable equation 2

**nameck** · March 26th 2010, 05:57 PM

By letting v = y/x, solve the following differential equation below by reducing them to equations where the variables are separable.

the solutions as the attachment..
the problem is, how to convert from..
dy/dx = (x^2 + y^2) / 2x^2

to...
xdv/dx = (v-1)^2 / 2
???

**dwsmith** · March 26th 2010, 06:10 PM

Instead start by saying y=vx; therefore, dy=xdv+vdx.

Now take your de and multiple by dx to both sides and you will be left with dy = all that stuff.

Subtract all that stuff to obtain:

dy-(blah)dx=0

From taking the derivative of y=vx, we know dy=xdv+vdx. Substitute that in where you see dy and sub in vx where ever there are xs.

**nameck** · March 26th 2010, 09:57 PM

still can't understand...

**dwsmith** · March 26th 2010, 10:08 PM

I typed up some of the steps in maple and converted to pdf.

You should be able to finish it from there.

**11rdc11** · March 26th 2010, 10:17 PM

Originally Posted by nameck

By letting v = y/x, solve the following differential equation below by reducing them to equations where the variables are separable.

the solutions as the attachment..
the problem is, how to convert from..
dy/dx = (x^2 + y^2) / 2x^2

to...
xdv/dx = (v-1)^2 / 2
???

$\frac{dy}{dx} = \frac{x^2+y^2}{2x^2} = \frac{x^2}{2x^2} +\frac{y^2}{2x^2}= \frac{1+v^2}{2}$

now since when sub in v we have to find

$v = \frac{y}{x}$

$vx =y$

$\frac{dv}{dx}x + v = \frac{dy}{dx}$

now put it all together

$\frac{dv}{dx}x + v = \frac{1+v^2}{2}$

$\frac{dv}{dx}x = \frac{1+v^2}{2} - v$

$\frac{dv}{dx}x = \frac{1+v^2}{2} - \frac{2v}{2}$

$\frac{dv}{dx}x = \frac{v^2-2v +1}{2}$

$\frac{dv}{dx}x = \frac{(v-1)^2}{2}$

$\int\frac{dv}{(v-1)^2} = \int\frac{dx}{2x}$

**mr fantastic** · March 26th 2010, 10:17 PM

Originally Posted by nameck

By letting v = y/x, solve the following differential equation below by reducing them to equations where the variables are separable.

the solutions as the attachment..
the problem is, how to convert from..
dy/dx = (x^2 + y^2) / 2x^2

to...
xdv/dx = (v-1)^2 / 2
???

$y = xv \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$ using the product rule.

And obviously $y = xv \Rightarrow \frac{x^2 + y^2}{2x^2} = \frac{x^2 + x^2 v^2}{2x^2} = ....$

**Prove It** · March 26th 2010, 10:23 PM

If $v = \frac{y}{x}$ then $y = vx$ .

If you take the derivative of both sides, you get

$\frac{d}{dx}(y) = \frac{d}{dx}(vx)$

$\frac{dy}{dx} = v\,\frac{d}{dx}(x) + x\,\frac{d}{dx}(v)$

$\frac{dy}{dx} = v + x\,\frac{dv}{dx}$ .

Putting this information into your DE:

$\frac{dy}{dx} = \frac{x^2 + y^2}{2x^2}$

$v + x\,\frac{dv}{dx} = \frac{x^2 + \left(vx\right)^2}{2x^2}$

$v + x\,\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x^2}$

$v + x\,\frac{dv}{dx} = \frac{1 + v^2}{2}$

$x\,\frac{dv}{dx} = \frac{1 + v^2}{2} - v$

$x\,\frac{dv}{dx} = \frac{v^2 - 2v + 1}{2}$

$x\,\frac{dv}{dx} = \frac{(v- 1)^2}{2}$

$\frac{2}{(v - 1)^2}\,\frac{dv}{dx} = \frac{1}{x}$

$\int{\frac{2}{(v - 1)^2}\,\frac{dv}{dx}\,dx} = \int{\frac{1}{x}\,dx}$

$\int{2(v - 1)^{-2}\,dv} = \int{\frac{1}{x}\,dx}$ .

You should be able to go from here.

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