Find the general solution

**alexmahone** · September 25th 2011, 07:44 AM

Find the general solution of the differential equation $3y+x^3y^4+3xy'=0$ .

$y'+\frac{y}{x}=\frac{-x^2y^4}{3}$

$xy'+y=\frac{-x^3y^4}{3}$

$\frac{d}{dx}(xy)=\frac{-x^3y^4}{3}$

$xy=\frac{-x^4y^4}{12}+C$

But the answer given in the book is $y(x)=x^{-1}(C+\ln x)^{-1/3}$ . Where have I gone wrong?

**chisigma** · September 25th 2011, 08:06 AM

Originally Posted by alexmahone

Find the general solution of the differential equation $3y+x^3y^4+3xy'=0$ .

$y'+\frac{y}{x}=\frac{-x^2y^4}{3}$

$xy'+y=\frac{-x^3y^4}{3}$

$\frac{d}{dx}(xy)=\frac{-x^3y^4}{3}$

$xy=\frac{-x^4y^4}{12}+C$

But the answer given in the book is $y(x)=x^{-1}(C+\ln x)^{-1/3}$ . Where have I gone wrong?

Setting $z=x\ y$ Your DE is...

$z^{'}= -\frac{z^{4}}{3 x}$ (1)

... and the integration is immediate...

$z= x y= -\frac{1}{(\ln x+c)^{\frac{1}{3}}}$ (2)

Kind regards

$\chi$ $\sigma$

**Prove It** · September 25th 2011, 08:07 AM

Originally Posted by alexmahone

Find the general solution of the differential equation $3y+x^3y^4+3xy'=0$ .

$y'+\frac{y}{x}=\frac{-x^2y^4}{3}$

$xy'+y=\frac{-x^3y^4}{3}$

$\frac{d}{dx}(xy)=\frac{-x^3y^4}{3}$

$xy=\frac{-x^4y^4}{12}+C$

But the answer given in the book is $y(x)=x^{-1}(C+\ln x)^{-1/3}$ . Where have I gone wrong?

You have gone wrong by treating y as a constant and not as a function of x.

You should recognise that this is actually a Bernoulli Equation.

**alexmahone** · September 25th 2011, 08:08 AM

Originally Posted by chisigma

Setting $z=x\ y$ Your DE is...

$z^{'}= -\frac{z^{4}}{3 x}$ (1)

... and the integration is immediate...

$z= x y= -\frac{1}{(\ln x+c)^{\frac{1}{3}}}$ (2)

Kind regards

$\chi$ $\sigma$

But what's wrong with my solution?

**alexmahone** · September 25th 2011, 08:11 AM

Originally Posted by Prove It

You have gone wrong by treating y as a constant and not as a function of x.

You should recognise that this is actually a Bernoulli Equation.

I know that this is a Bernoulli equation; that is why I multiplied by the integrating factor x. Which step of my solution is actually invalid?

**Prove It** · September 25th 2011, 08:13 AM

Originally Posted by alexmahone

I know that this is a Bernoulli equation; that is why I multiplied by the integrating factor x. Which step of my solution is actually invalid?

Which steps are invalid?

1. Not making an appropriate substitution.

2. Performing integration treating a function of the independent variable like a constant...

**chisigma** · September 25th 2011, 08:15 AM

Originally Posted by alexmahone

But what's wrong with my solution?

You arrived [correctly...] to write...

$\frac{d}{dx} (x y)= - \frac{x^{3}y^{4}}{3}$ (1)

... and then You [incorrectly...] operated as in case of separate variables... but in (1) x and y aren't separate...

Kind regards

$\chi$ $\sigma$

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