1. bernoulli

2x dy/dx + 3y + (x+1)y^3 = 0

i try to get it in the form
dy/dx + fxy = gxy^n but it isnt working
im geting loads of denominators which cant be solved
cheers

2. Have you used the standard substitution $\displaystyle v=y^{-2}$ ?. In such case, what do you obtain?

Fernando Revilla

3. i made that substitution and did the dividing through to get

dv/dx - 3v/x = (x+1)/x

is this correct?
which is linear in v but again i dont know what to do with it

4. i cant seperate the variables with the x on the botom

5. Now multiply both sides by the integrating factor $\displaystyle \displaystyle e^{\int{-\frac{3}{x}\,dx}} = e^{-3\ln{x}} = e^{\ln{(x^{-3})}} = x^{-3}$ to turn the left hand side into a product-rule expansion, which you can write as a single derivative.

6. Originally Posted by mathcore
i made that substitution and did the dividing through to get dv/dx - 3v/x = (x+1)/x is this correct?

Yes, it is. Now solve the homogeneous equation $\displaystyle dv/dx-3v/x=0$ or equivalently $\displaystyle dv/v-3dx/x=0$ .

Fernando Revilla

7. Originally Posted by FernandoRevilla
Yes, it is. Now solve the homogeneous equation $\displaystyle dv/dx-3v/x=0$ or equivalently $\displaystyle dv/v-3dx/x=0$ .

Fernando Revilla
Except $\displaystyle \displaystyle \frac{dv}{dx} - \frac{3v}{x} = \frac{x + 1}{x}$ is not a homogeneous equation, and is not separable.

Since it's first-order linear, the Integrating Factor method is needed...

8. it cannot be homogeneous. this example is not homogenous as it doesnt require knowledge of homogeneous d.es?

yes i'm going to go with the i.f

Originally Posted by Prove It
Now multiply both sides by the integrating factor $\displaystyle \displaystyle e^{\int{-\frac{3}{x}\,dx}} = e^{-3\ln{x}} = e^{\ln{(x^{-3})}} = x^{-3}$ to turn the left hand side into a product-rule expansion, which you can write as a single derivative.
my log rules need some sharpening up, i didn't realise it came to x^-3, i ended up with just 3 which i knew had to be wrong as it was useless.

how do you integrate -3/x? do you just take the -3 out and integrate 1/x?

i laughed so bad when i saw the product rule thing. this is a really nice question.

the question is what can be done about the RHS because i dont know what to do with it at all

9. You should have gotten $\displaystyle \displaystyle \frac{d}{dx}(\textrm{something}) = \frac{x+ 1}{x}$, which is equivalent to

$\displaystyle \displaystyle \textrm{something} = \int{\frac{x + 1}{x}\,dx}$.

Also note that $\displaystyle \displaystyle \frac{x + 1}{x} = 1 + \frac{1}{x}$.

And to answer your other question, $\displaystyle \displaystyle \int{-\frac{3}{x}\,dx} = -3\int{\frac{1}{x}\,dx}$, like you suspect.

10. i got d/dx x^-3.v = int(x+1)x^-4

the rhs is multiplied by the integrating factor too... i dont know how to integrate that

11. True. Notice that $\displaystyle \displaystyle \frac{x + 1}{x^4} = x^{-3} + x^{-4}$. Now integrate each term...

12. thanks i think i get it!

Wait .... what i dont get is how to change x+1/x into that easy form?

13. $\displaystyle \displaystyle \frac{x + 1}{x} = \frac{x}{x} + \frac{1}{x} = 1 + \frac{1}{x}$...

14. ok thanks now i realise i have to sub back in v=y^-2 and i am stuck because i am left with my answer in terms of y^-2.

i get x^-3.y^-2 = -1/2.x^-2 -1/3.x^-3 + constant and i am lost.

15. Step 1: Solve for $\displaystyle \displaystyle y^{-2}$.

Step 2: Take both sides to the power of $\displaystyle \displaystyle -\frac{1}{2}$.

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