1. Bernoulli equation

Hello,
I want to solve this bernoulli differential equation.But my answer is different.

$y'+2xy=xy^2$

Multiplying both sides of equation by $y^{-2}$ and -1, we get

$-y^{-2}y'-xy^{-1}=-x$ Let $z=y^{-1}$ Then,the equation became z'-xz=-x. Now the equation is linear in z.Using P(x)=-x produces $\int P(x)dx=\int -xdx=\frac{-x^2}{2}$ which implies that $e^{-x^2/2}$is an integrating factor.Multiplying the linear equation by this factor we get

$z'e^{-x^2/2}-xze^{-x^2/2}=-xe^{-x^2/2}$

$\frac{d[ze^{-x^2/2}]}{dx}=-xe^{-x^2/2}$

$ze^{-x^2/2}=\int -xe^{-x^2/2}dx$

$ze^{-x^2/2}=e^{-x^2/2}+c$
$y^{-1}=1+ce^{x^2/2}$
$y=\frac{1}{1+ce^{x^2/2}}$

But answer is $\frac{-2}{Ce^{x^2}-1}$ Where is i am wrong in my answer, would any forumite explain me?

2. Re: Bernoulli equation

Hey Vinod.

Hint - Think about when you integrate e^[cu] du and the constant c.

3. Re: Bernoulli equation Originally Posted by chiro Hey Vinod.

Hint - Think about when you integrate e^[cu] du and the constant c.
Hello chiro,
Integrating $e^{cu}du$, we get $e^{cu}/c$. that's it.

4. Re: Bernoulli equation Originally Posted by chiro Hey Vinod.

Hint - Think about when you integrate e^[cu] du and the constant c.
Hello,
After rectification of my wrong answer in post#1, i got the answer $\frac{2}{1+2ce^{x^2}}$ but it is also different.

5. Re: Bernoulli equation

2 times some arbitrary constant is still just an arbitrary constant, right?

6. Re: Bernoulli equation Originally Posted by MarkFL 2 times some arbitrary constant is still just an arbitrary constant, right?
Hello, Yes, You are right.Answer i got is not different from the answer provided.