# [SOLVED] differential equation

• February 1st 2009, 11:17 AM
rajr
[SOLVED] differential equation
Determine the general solution of the differential equation

y′ = (y/2x) − (xy)^3 .

what method should i use for this
• February 1st 2009, 11:24 AM
Chris L T521
Quote:

Originally Posted by rajr
c)
Determine the general solution of the differential equation

y
= (y/2x) (xy)^3 .

what method should i use for this

Rearranging, you get $\frac{\,dy}{\,dx}-\frac{1}{2x}y=-x^3y^3$

This can be recognized as a Bernoulli Equation.

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

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

Making the appropriate substitutions, we end up with:

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

Can you take it from here?
• February 1st 2009, 11:41 AM
rajr
thnk u

is the integrating factor is ln¦x¦
• February 1st 2009, 11:41 AM
Chris L T521
Quote:

Originally Posted by rajr
thnk u

is the integrating factor is ln¦x¦

The integrating factor is ${\color{red}e}^{\ln x}=x$
• February 1st 2009, 11:53 AM
rajr

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
• February 1st 2009, 11:59 AM
Chris L T521
Quote:

Originally Posted by rajr

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

You're correct! (Clapping)
• February 1st 2009, 12:02 PM
Jhevon
Quote:

Originally Posted by rajr

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

yes, if by 2x^4/5 you mean $\frac {2x^4}5$
• February 1st 2009, 12:04 PM
rajr
Quote:

Originally Posted by Chris L T521
You're correct! (Clapping)

yes, if by 2x^4/5 you mean $\frac {2x^4}5$