ODE's

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Hello

Joined Dec 18, 2008
82
Using the substitution y = 1/z, find the solution of the following differential equation:

dy/dx = y + x(y^2)



This is what i've done so far:

dy/dx = (dy/dz).(dz/dx)

Not sure what to do next or if it is correct!

Any help would be greatly appreciated.
 
Last edited:

steveb

Joined Jul 3, 2008
2,436
This is what i've done so far:

dy/dx = (dy/dz).(dz/dx)

Not sure what to do next or if it is correct!

Any help would be greatly appreciated.
That is a good start. Now complete the substitution in the entire equation. The dy/dz can be easily identified from y=1/z (i.e. dy/dz=-1/z^2)

Once you make the substitution you will see that the final equation is in a very simple form.

dz/dx=-z-x

This is a simple first order system with a negative ramp (-x) as the input function.
 

Ratch

Joined Mar 20, 2007
1,070
Hello,

Using the substitution y = 1/z, find the solution of the following differential equation:

dy/dx = y + x(y^2)
dy/dx = (-1/z^2)dz/dx

so (-1/z^2)dz/dx = 1/z + x/z^2 ==> -dz/dx = z + x ==> -dz = zdx + xdx

dz + zdx = -xdx ==> (e^x)dz + z(e^x)dx = -x(e^x)dx ==> d((e^x)z) = d((e^x-x(e^x)) ==> (e^x)z = (e^x-x(e^x) +C ==> z = 1-x+C/e^x
==> 1/y = 1-x+C/e^x ==> y = 1/(1-x+C/e^x)

Ratch
 
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