1. ## (Bernoulli's Eq)

Bernoullis' Equation is given by: $\displaystyle \frac{dy}{dt} + P(t)y = f(t)y^n$ (1) where $n \in \mathbb{R}$. Solve this Bernoulli equation:

$\displaystyle t^2\frac{dy}{dt} + 2ty-y^3 = 0$

NOTE: You may want to use the substitution $u = y^{1-n}$ which will reduce any nonlinear equation of the form above in (1), where $n \ne 0$ and $n \ne 1$

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So this looks like an integrating factor problem, but I get stuck when doing it. And the substitution thing gets confusing.

2. Originally Posted by seerTneerGevoLI
$\displaystyle t^2\frac{dy}{dt} + 2ty-y^3 = 0$
$t^2 y' + 2ty - y^3 = 0 \iff \frac{{y'}}
{{y^3 }}t^2 + \frac{{2t}}
{{y^2 }} = 1$

Define $u=\frac1{y^2}.$ After that you can turn the Bernoulli equation into a Linear one.

3. Originally Posted by Krizalid
$t^2 y' + 2ty - y^3 = 0 \iff \frac{{y'}}
{{y^3 }}t^2 + \frac{{2t}}
{{y^2 }} = 1$

Define $u=\frac1{y^2}.$ After that you can turn the Bernoulli equation into a Linear one.
Hmm, so is this substitution, then:

$\displaystyle \frac{y'ut^2}{y} + 2tu = 1$

I don't see how that helps much. Now we just have introduced one extra variable. Can't I just see that 2/y^2 is an integrating factor and try solve this then?

4. Careful...

Since $u=\frac1{y^2}\implies u'=-\frac{2y'}{y^3}$

Now substitute and yields $-\frac12u't^2+2tu=1$

Can you see that this new equation is linear?

5. Originally Posted by Krizalid
Careful...

Since $u=\frac1{y^2}\implies u'=-\frac{2y'}{y^3}$

Now substitute and yields $-\frac12u't^2+2tu=1$

Can you see that this new equation is linear?
Ah, thank you ever so much. I think I'm done. Let me know if you see errors:

$\displaystyle t^2 y' + 2ty - y^3 = 0 \iff \frac{{y'}}{{y^3 }}t^2 + \frac{{2t}}{{y^2 }} = 1$

$u=\frac1{y^2}\implies u'=-\frac{2y'}{y^3}$

$-\frac{1}{2}u't^2+2tu=1$

$\displaystyle u' + \frac{2u}{t} = \frac{-2}{t^2}$

Our integrating factor, M(t):

$\displaystyle M(t) = e^{\int{\frac{2}{t}\,dt}} = e^{2\ln{|t|}} = t^2$

Multiply each term by M(t):

$\displaystyle t^2u' + \frac{2u}{t}t^2 = -\frac{2}{t^2}t^2 \implies t^2u' + 2ut = -2$

Use product rule in reverse:

$\displaystyle \left(t^2u\right)' = -2$

Integrate both sides:

$t^2u = -2t + C$

Substitute back in for u:

$t^2\frac{1}{y^2} = -2t + C \implies \frac{t^2}{y^2} = -2t + C\implies t^2 = -2ty^2 + y^2C$
$\implies t^2 = (-2t + C)y^2 \implies \frac{t^2}{-2t+C} = y^2$
$\implies \boxed{y = \sqrt{\frac{t^2}{-2t+C}}}$

Thanks for checking my work.

6. Originally Posted by seerTneerGevoLI
Ah, thank you ever so much. I think I'm done. Let me know if you see errors:

$\displaystyle t^2 y' + 2ty - y^3 = 0 \iff \frac{{y'}}{{y^3 }}t^2 + \frac{{2t}}{{y^2 }} = 1$

$u=\frac1{y^2}\implies u'=-\frac{2y'}{y^3}$

$-\frac{1}{2}u't^2+2tu=1$

$\displaystyle u' { \color{red}+ \frac{2u}{t}} = \frac{-2}{t^2}$

7. Originally Posted by Jhevon
How? $\displaystyle -\frac{1}{2}u't^2 + 2tu = 1 \implies u't^2 + 2tu = -2 \implies u' + \frac{2u}{t} = \frac{-2}{t^2}$

EDIT: !!!! I didn't multiply the 2tu by (-2) grrr, what a waste of my time.

8. Originally Posted by seerTneerGevoLI
How? $\displaystyle -\frac{1}{2}u't^2 + 2tu = 1 \implies u't^2 + 2tu = -2 \implies u' + \frac{2u}{t} = \frac{-2}{t^2}$
you should have caught yourself by doing it again. to get rid of the $- \frac 12$ in front, you multiplied through by -2, correct? what is $-2 \left( 2tu\right)$?

EDIT: Ok, you caught yourself, good job. well, it wasn't a waste of time. you'll be more careful with such things from now on, and it will save you time in the future. we tend to make less mistakes if it hurts to make them

9. Alright, Jhevon. How does this look?

$\displaystyle t^2 y' + 2ty - y^3 = 0 \iff \frac{{y'}}{{y^3 }}t^2 + \frac{{2t}}{{y^2 }} = 1$

$u=\frac1{y^2}\implies u'=-\frac{2y'}{y^3}$

$-\frac{1}{2}u't^2+2tu=1$

$\displaystyle u' - \frac{4u}{t} = \frac{-2}{t^2}$

Integrating factor, M(t) is:

$\displaystyle M(t) = e^{\int{-\frac{4}{t}\,dt}} = e^{-4\ln{|t|}} = \frac{1}{t^4}$

Multiply each term by M(t)

$\displaystyle \frac{1}{t^4}u' - \frac{4u}{t}\cdot \frac{1}{t^4} = -\frac{2}{t^2}\cdot \frac{1}{t^4} \implies \frac{u'}{t^4}-\frac{4u}{t^5} = \frac{-2}{t^6}$

Use product rule in reverse:

$\displaystyle \left(\frac{u}{t^4}\right)' = \frac{-2}{t^6}$

Take integral

$\displaystyle \frac{u}{t^4} = \frac{2}{5t^5} + C$

Subt. back in for u

$\frac{\frac{1}{y^2}}{t^4} = \frac{2}{5t^5} + C \implies \frac{1}{t^4y^2} = \frac{2}{5t^5} + C$

Solve for y:

$\displaystyle \frac{1}{t^4y^2} = \frac{2}{5t^5} + C \implies t^4y^2 = \frac{2}{5t^5} + \frac{1}{C} \implies y^2 = \frac{2}{5t^6} + \frac{1}{t^4C}$

Thus,

$y = \sqrt{\frac{2}{5t^6} + \frac{1}{t^4C}}$

QUESTION ON THIS LAST PART: is it y = +\- sqrt(...) or do we not have to worry about the negative.

Thanks

10. Originally Posted by seerTneerGevoLI
Alright, Jhevon. How does this look?

$\displaystyle t^2 y' + 2ty - y^3 = 0 \iff \frac{{y'}}{{y^3 }}t^2 + \frac{{2t}}{{y^2 }} = 1$

$u=\frac1{y^2}\implies u'=-\frac{2y'}{y^3}$

$-\frac{1}{2}u't^2+2tu=1$

$\displaystyle u' - \frac{4u}{t} = \frac{-2}{t^2}$

Integrating factor, M(t) is:

$\displaystyle M(t) = e^{\int{-\frac{4}{t}\,dt}} = e^{-4\ln{|t|}} = \frac{1}{t^4}$

Multiply each term by M(t)

$\displaystyle \frac{1}{t^4}u' - \frac{4u}{t}\cdot \frac{1}{t^4} = -\frac{2}{t^2}\cdot \frac{1}{t^4} \implies \frac{u'}{t^4}-\frac{4u}{t^5} = \frac{-2}{t^6}$

Use product rule in reverse:

$\displaystyle \left(\frac{u}{t^4}\right)' = \frac{-2}{t^6}$

Take integral

$\displaystyle \frac{u}{t^4} = \frac{2}{5t^5} + C$

Subt. back in for u

$\frac{\frac{1}{y^2}}{t^4} = \frac{2}{5t^5} + C \implies \frac{1}{t^4y^2} = \frac{2}{5t^5} + C$

Solve for y:

$\displaystyle \frac{1}{t^4y^2} = \frac{2}{5t^5} + C \implies {\color{red}t^4y^2 = \frac{2}{5t^5} + \frac{1}{C} \implies y^2 = \frac{2}{5t^6} + \frac{1}{t^4C}}$
you made another mistake here. you inverted both sides incorrectly. combine the fractions on the right first, then flip both sides. (by the way, i would multiply through by $t^4$ just before that)

11. Third time is the charm. Luckily this time it's not too far up.

$\displaystyle \frac{1}{t^4y^2} = \frac{2}{5t^5} + C \implies \frac{1}{y^2} = \frac{2}{5t} + t^4C \implies \frac{1}{y^2} = \frac{5ty^4C + 2}{5t} \implies y^2 = \frac{5t}{5ty^4C + 2}$

Hence, the general solution is:

$y = \sqrt{\frac{5t}{5ty^4C + 2}}$

Same question applies whether I need +/- or not. I'm hoping this is right now.

12. Originally Posted by seerTneerGevoLI
Third time is the charm. Luckily this time it's not too far up.

$\displaystyle \frac{1}{t^4y^2} = \frac{2}{5t^5} + C \implies \frac{1}{y^2} = \frac{2}{5t} + t^4C \implies \frac{1}{y^2} = \frac{5ty^4C + 2}{5t} \implies y^2 = \frac{5t}{5ty^4C + 2}$

Hence, the general solution is:

$y = \sqrt{\frac{5t}{5ty^4C + 2}}$

Same question applies whether I need +/- or not. I'm hoping this is right now.
there should be no y on the right side of the equation. check to see where you made a mistake. you answer should be: $y(t) = \pm \sqrt{\frac {5t}{5Ct^5 + 2}}$

13. Originally Posted by seerTneerGevoLI
Same question applies whether I need +/- or not. I'm hoping this is right now.
yes, we need the +/- here. we have no basis on which to choose the positive over the negative. if we were given initial conditions or some other conditions to fulfill that would cause a negative or a positive answer to be impossible, then we would drop one of the signs