Bernoulli - Find the general solution

EmmanuelD
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Bernoulli -- Find the general solution

Homework Statement



Find the general solution of:

y'+xy=xy^3


Homework Equations



Bernoulli's Equation


The Attempt at a Solution



y'+xy=xy^3

(y^-3)y'+x(y^-2)=x

Let v=(y^-2), thus v'=((-2y^-3)y'

Then,

-v'/2+xv=x

Multiply through by (-2)

v'-2xv=-2x

Now it's in first-order linear so I multiply by e^int(-2x)dx = e^(-x^2)

(e^(-x^2))v'-2xe^(-x^2)=2xe^(-x^2)

This is where I'm getting stuck :(

BECAUSE: the integral of e^(-x^2) is the "error function?"

According to Wolfram, integral e^(-x^2) dx = 1/2 sqrt(pi) erf(x)+constant

----

Anyone get anything different or know where I might have messed up?

THANK YOU!
 
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EmmanuelD said:

Homework Statement



Find the general solution of:

y'+xy=xy^3


Homework Equations



Bernoulli's Equation


The Attempt at a Solution



y'+xy=xy^3

(y^-3)y'+x(y^-2)=x

Let v=(y^-2), thus v'=((-2y^-3)y'

Then,

-v'/2+xv=x

Multiply through by (-2)

v'-2xv=-2x

Now it's in first-order linear so I multiply by e^int(-2x)dx = e^(-x^2)

(e^(-x^2))v'-2xe^(-x^2)=2xe^(-x^2)

This is where I'm getting stuck :(

BECAUSE: the integral of e^(-x^2) is the "error function?"

According to Wolfram, integral e^(-x^2) dx = 1/2 sqrt(pi) erf(x)+constant

----

Anyone get anything different or know where I might have messed up?

THANK YOU!

GOT IT! Sorry, I don't know if there's an option for deleting a topic. My apologies.

Thanks, either way :)
 


Should have used separation of variables.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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