# Find the general solution

• Sep 25th 2011, 07:44 AM
alexmahone
Find the general solution
Find the general solution of the differential equation $\displaystyle 3y+x^3y^4+3xy'=0$.

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

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

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

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

But the answer given in the book is $\displaystyle y(x)=x^{-1}(C+\ln x)^{-1/3}$. Where have I gone wrong?
• Sep 25th 2011, 08:06 AM
chisigma
Re: Find the general solution
Quote:

Originally Posted by alexmahone
Find the general solution of the differential equation $\displaystyle 3y+x^3y^4+3xy'=0$.

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

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

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

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

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

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

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

... and the integration is immediate...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 25th 2011, 08:07 AM
Prove It
Re: Find the general solution
Quote:

Originally Posted by alexmahone
Find the general solution of the differential equation $\displaystyle 3y+x^3y^4+3xy'=0$.

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

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

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

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

But the answer given in the book is $\displaystyle y(x)=x^{-1}(C+\ln x)^{-1/3}$. 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.
• Sep 25th 2011, 08:08 AM
alexmahone
Re: Find the general solution
Quote:

Originally Posted by chisigma
Setting $\displaystyle z=x\ y$ Your DE is...

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

... and the integration is immediate...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

But what's wrong with my solution?
• Sep 25th 2011, 08:11 AM
alexmahone
Re: Find the general solution
Quote:

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?
• Sep 25th 2011, 08:13 AM
Prove It
Re: Find the general solution
Quote:

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...
• Sep 25th 2011, 08:15 AM
chisigma
Re: Find the general solution
Quote:

Originally Posted by alexmahone
But what's wrong with my solution?

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$