Get Help with Initial Value Problem: Solving xy' + 6y = 3xy^3 Equation

[Pengwuino]

I have a initial value problem that I have no idea how to start! Ahhh :(

The equation is:

$$xy' + 6y = 3xy^3$$

How do I start? I have no idea! ahhh I don't like DE's!

 

[NINHARDCOREFAN]

Seperation of vaiables.

 

[d_leet]

I have a initial value problem that I have no idea how to start! Ahhh :(

The equation is:

$$xy' + 6y = 3xy^3$$

How do I start? I have no idea! ahhh I don't like DE's!

I can give you a link that should help, but I can't explain it very well because I'm still learning this too. Start by dividing everything by x then check out this link http://en.wikipedia.org/wiki/Bernoulli_differential_equation

 

[d_leet]

Seperation of vaiables.

I'm pretty sure that isn't going to work on this equation.

OK I was wrong it'll work it just wasn't obvious to me at first sorry.

 

[Pengwuino]

So all x and y variables will be seperated? I can't seem to get them to seperate...

 

[d_leet]

So all x and y variables will be seperated? I can't seem to get them to seperate...

Try dividing everything by y and then isolating the y' term that might make it more obvious.

 

[Pengwuino]

Can i even do this?

$$\begin{array}{l} \frac{{xy' + 6y}}{y} = \frac{{3xy^3 }}{y} \\ \frac{{xy'}}{y} + 6 = 3xy^2 \\ \frac{{xy'}}{y} = 3xy^2 \\ xy' = 3xy^3 \\ y' = 3y^3 \\ \frac{{dy}}{{dx}} = 3y^3 \\ \frac{{dy}}{{3y^3 dx}} = 1 \\ \frac{1}{{3y^3 }}dy = dx \\ \end{array}$$

If that's the case, I guess i know exactly what to do next!

 

[d_leet]

Can i even do this?

$$\begin{array}{l} \frac{{xy' + 6y}}{y} = \frac{{3xy^3 }}{y} \\ \frac{{xy'}}{y} + 6 = 3xy^2 \\ \frac{{xy'}}{y} = 3xy^2 \\ xy' = 3xy^3 \\ y' = 3y^3 \\ \frac{{dy}}{{dx}} = 3y^3 \\ \frac{{dy}}{{3y^3 dx}} = 1 \\ \frac{1}{{3y^3 }}dy = dx \\ \end{array}$$

If that's the case, I guess i know exactly what to do next!

You dropped a 6 out somewhere in teh middle and I realized that I screwed up when I tried to separate the variables so I guess it isn't separable after all, try the method in the link I posted above.

 

[Pengwuino]

Wow i still can't get this...

 

[Cyrus]

$$y' + \frac{6y}{x} = 3 y^3$$

divide by $$y^3$$

$$y'y^{-3} + \frac{6y^-2}{x} = 3$$

Make a change of variables:

$$w = \frac{1}{y^2}$$

and

$$w' = \frac{1-3}{y^3} y' = \frac{-2}{y^3} y'$$



This leads to:

$$- \frac{w'}{2} + \frac{6w}{x} = 3$$

we-write this in standard form:

$$\frac{dw}{dx} - \frac{12w}{x} = -6$$

solve using an integrating factor of:

$$M(x) = e^{\int P(x)dx } = e^{\int \frac{-12}{x}dx} = e^{-12ln(x)} = x^{-12}$$

multiply both sides by M(x):

$$\frac{w'}{x^{-12}} - \frac{12w}{x^{-13}} = -\frac{6}{x^{-12}}$$

but that is equal to:

$$\frac { d( \frac{w} {x^{-12}})} {dx} = -\frac{6}{x^{-12}}$$

integrate both sides:

$$\frac{w} {x^{-12}} = \frac {6}{11} x^{-11} + c$$

Solve in terms of w:

$$w = \frac{6}{11} x + cx^{12}$$

Now plug back in for w:

$$\frac{1}{y^2} = \frac{6}{11} x + cx^{12}$$

 

[Pengwuino]

Now i got to figure out how to even use Bernoulli's equation.

 

[Cyrus]

Ok, I hope the anwser is correct. I probably have some mistakes in there. I am REALLY TIRED right now, and I HAVE to go to sleep. Please check it over someone:

The solution is:

$$\frac{1}{y^2} = \frac{6}{11} x + cx^{12}$$