# Solve in two different ways

• September 25th 2011, 11:35 PM
alexmahone
Solve in two different ways
Solve in two different ways $\frac{dy}{dx}=xy^3-xy$.

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Separable:

$\frac{dy}{dx}=x(y^3-y)$

$\frac{dy}{y^3-y}=xdx$

$\frac{dy}{y(y+1)(y-1)}=xdx$

Let $\frac{1}{y(y+1)(y-1)}=\frac{A}{y}+\frac{B}{y+1}+\frac{C}{y-1}$

$1=A(0+1)(0-1)$

$A=-1$

$1=B(-1)(-1-1)$

$B=\frac{1}{2}$

$1=C(1+1)$

$C=\frac{1}{2}$

$\int \left[\frac{-1}{y}+\frac{1}{2(y+1)}+\frac{1}{2(y-1)}\right]dy=\int xdx$

$-\ln y+\frac{1}{2}\ln (y+1)+\frac{1}{2}\ln (y-1)=\frac{x^2}{2}+C'$

$-2\ln y+\ln (y+1)+\ln (y-1)=x^2+C$, where $C=2C'$.

$\ln \frac{(y+1)(y-1)}{y^2}=x^2+C$

Bernoulli:

$\frac{dy}{dx}+xy=xy^3$

Let $v=y^{-2}$.

$y=v^{-1/2}$

$\frac{dy}{dx}=-\frac{1}{2}v^{-3/2}\frac{dv}{dx}$

$-\frac{1}{2}v^{-3/2}\frac{dv}{dx}+xv^{-1/2}=xv^{-3/2}$

Multiplying throughout by $-2v^{3/2}$,

$\frac{dv}{dx}-2xv=-2x$

The integrating factor is $e^{\int -2xdx}=e^{-x^2}$.

$e^{-x^2}\frac{dv}{dx}-2xe^{-x^2}v=-2xe^{-x^2}$

$\frac{d}{dx}(ve^{-x^2})=-2xe^{-x^2}$

$ve^{-x^2}=e^{-x^2}+C$

$y^{-2}e^{-x^2}=e^{-x^2}+C$

$y^{-2}=Ce^{x^2}+1$

$y=(Ce^{x^2}+1)^{-1/2}$

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Where have I gone wrong?
• September 25th 2011, 11:41 PM
Prove It
Re: Solve in two different ways
Quote:

Originally Posted by alexmahone
Solve in two different ways $\frac{dy}{dx}=xy^3-xy$.

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Separable:

$\frac{dy}{dx}=x(y^3-y)$

$\frac{dy}{y^3-y}=xdx$

$\frac{dy}{y(y+1)(y-1)}=xdx$

Let $\frac{1}{y(y+1)(y-1)}=\frac{A}{y}+\frac{B}{y+1}+\frac{C}{y-1}$

$1=A(0+1)(0-1)$

$A=-1$

$1=B(-1)(-1-1)$

$B=\frac{1}{2}$

$1=C(1+1)$

$C=\frac{1}{2}$

$\int \left[\frac{-1}{y}+\frac{1}{2(y+1)}+\frac{1}{2(y-1)}\right]dy=\int xdx$

$-\ln y+\frac{1}{2}\ln (y+1)+\frac{1}{2}\ln (y-1)=\frac{x^2}{2}+C'$

$-2\ln y+\ln (y+1)+\ln (y-1)=x^2+C$, where $C=2C'$.

$\ln \frac{(y+1)(y-1)}{y^2}=x^2+C$

Bernoulli:

$\frac{dy}{dx}+xy=xy^3$

Let $v=y^{-2}$.

$y=v^{-1/2}$

$\frac{dy}{dx}=-\frac{1}{2}v^{-3/2}\frac{dv}{dx}$

$-\frac{1}{2}v^{-3/2}\frac{dv}{dx}+xv^{-1/2}=xv^{-3/2}$

Multiplying throughout by $-2v^{3/2}$,

$\frac{dv}{dx}-2xv=-2x$

The integrating factor is $e^{\int -2xdx}=e^{-x^2}$.

$e^{-x^2}\frac{dv}{dx}-2xe^{-x^2}v=-2xe^{-x^2}$

$\frac{d}{dx}(ve^{-x^2})=-2xe^{-x^2}$

$ve^{-x^2}=e^{-x^2}+C$

$y^{-2}e^{-x^2}=e^{-x^2}+C$

$y^{-2}=Ce^{x^2}+1$

$y=(Ce^{x^2}+1)^{-1/2}$

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Where have I gone wrong?

What makes you think there's something wrong with either of them (except for leaving off the modulus signs in your logarithms)? If you try to write your separable equation as y in terms of x, you should get something that is correct up to an integration constant.
• September 25th 2011, 11:50 PM
alexmahone
Re: Solve in two different ways
Quote:

Originally Posted by Prove It
What makes you think there's something wrong with either of them (except for leaving off the modulus signs in your logarithms)? If you try to write your separable equation as y in terms of x, you should get something that is correct up to an integration constant.

$\ln \left|\frac{(y+1)(y-1)}{y^2}\right|=x^2+C$

$\frac{(y+1)(y-1)}{y^2}=\pm e^{(x^2+C)}=\pm e^Ce^{x^2}$

$\frac{(y+1)(y-1)}{y^2}=C'e^{x^2}$, where $C'=\pm e^C$

$y^2-1=C'e^{x^2}y^2$

$y^2(1-C'e^{x^2})=1$

$y^2=(1-C'e^{x^2})^{-1}$

$y=(1-C'e^{x^2})^{-1/2}$, where $C'$ is the negative of the constant in the Bernoulli method.