1. ## [SOLVED] differential equation

Determine the general solution of the differential equation

y′ = (y/2x) − (xy)^3 .

what method should i use for this

2. 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?

3. thnk u

is the integrating factor is ln¦x¦

4. Originally Posted by rajr
thnk u

is the integrating factor is ln¦x¦
The integrating factor is ${\color{red}e}^{\ln x}=x$

5. 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

6. 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!

7. 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 $\frac {2x^4}5$

8. Originally Posted by Chris L T521
You're correct!
yes, if by 2x^4/5 you mean $\frac {2x^4}5$