Determine the general solution of the differential equation

y′ = (y/2x) − (xy)^3 .

what method should i use for this

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- Feb 1st 2009, 11:17 AMrajr[SOLVED] differential equation
Determine the general solution of the differential equation

y′ = (y/2x) − (xy)^3 .

what method should i use for this - Feb 1st 2009, 11:24 AMChris L T521
Rearranging, you get $\displaystyle \frac{\,dy}{\,dx}-\frac{1}{2x}y=-x^3y^3$

This can be recognized as a Bernoulli Equation.

Let $\displaystyle z=y^{1-3}=y^{-2}\implies z^{-\frac{1}{2}}=y$

Thus, $\displaystyle \frac{\,dy}{\,dx}=\frac{\,dy}{\,dz}\cdot\frac{\,dz }{\,dx}=-\tfrac{1}{2}z^{-\frac{3}{2}}\frac{\,dz}{\,dx}$

Making the appropriate substitutions, we end up with:

$\displaystyle -\tfrac{1}{2}z^{-\frac{3}{2}}\frac{\,dz}{\,dx}-\frac{1}{2x}z^{-\frac{1}{2}}=-x^{3}z^{-\frac{3}{2}}$ which becomes the linear equation $\displaystyle \frac{\,dz}{\,dx}+\frac{1}{x}z=2x^3$

Can you take it from here? - Feb 1st 2009, 11:41 AMrajr
thnk u

is the integrating factor is ln¦x¦ - Feb 1st 2009, 11:41 AMChris L T521
- Feb 1st 2009, 11:53 AMrajr
hi thx for ur reply

i got z to be (2x^4/5) + (c/x)

i got the final answer to be 1/[sqrt{(2x^4/5) + (c/x)}]

is this right - Feb 1st 2009, 11:59 AMChris L T521
- Feb 1st 2009, 12:02 PMJhevon
- Feb 1st 2009, 12:04 PMrajr
- Feb 1st 2009, 12:10 PMrajr