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}<br /> \frac{{xy&#039; + 6y}}{y} = \frac{{3xy^3 }}{y} \\ <br /> \frac{{xy&#039;}}{y} + 6 = 3xy^2 \\ <br /> \frac{{xy&#039;}}{y} = 3xy^2 \\ <br /> xy&#039; = 3xy^3 \\ <br /> y&#039; = 3y^3 \\ <br /> \frac{{dy}}{{dx}} = 3y^3 \\ <br /> \frac{{dy}}{{3y^3 dx}} = 1 \\ <br /> \frac{1}{{3y^3 }}dy = dx \\ <br /> \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}<br /> \frac{{xy&#039; + 6y}}{y} = \frac{{3xy^3 }}{y} \\ <br /> \frac{{xy&#039;}}{y} + 6 = 3xy^2 \\ <br /> \frac{{xy&#039;}}{y} = 3xy^2 \\ <br /> xy&#039; = 3xy^3 \\ <br /> y&#039; = 3y^3 \\ <br /> \frac{{dy}}{{dx}} = 3y^3 \\ <br /> \frac{{dy}}{{3y^3 dx}} = 1 \\ <br /> \frac{1}{{3y^3 }}dy = dx \\ <br /> \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&#039; + \frac{6y}{x} = 3 y^3

divide by y^3

y&#039;y^{-3} + \frac{6y^-2}{x} = 3

Make a change of variables:

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

and

w&#039; = \frac{1-3}{y^3} y&#039; = \frac{-2}{y^3} y&#039;



This leads to:

- \frac{w&#039;}{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&#039;}{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}