Bernoulli - Find the general solution

EmmanuelD
Messages
10
Reaction score
0
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!
 
Physics news on Phys.org


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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top