transforming from a non-exact to an exact linear differential equation

The question asks me to solve the following equation.

3(x^2 + y^2)dx+x(x^2 + 3y^2 + 6y)dy = 0

I verified that this form is not exact by differentiating M with respect to y and N with respect to x and saw they were not equal. the problem now is to find a suitable u(x,y) such that when I multiply it by the equation, the equation becomes exact.

one of the ways I tried to transform the equation was to see if it was linear. what I got was

x' - x (3y^2 + 6y)/(x^2+y^2) = -x^3 / (x^2 + y^2) ...

hints would be greatly appreciated. also any good site on typing with Latex. all I get are 'unknown errors'

Re: transforming from a non-exact to an exact linear differential equation

Try $\displaystyle u(x,y)=h(x^{m}y^{n}).$

It may be more algebra, calculus, and differential equations than other methods, but it takes out some of the guesswork. What do you get with this?

Re: transforming from a non-exact to an exact linear differential equation

To add to Ackbeet's suggestion, try $\displaystyle u(x,y)=h(y) $ only!

Re: transforming from a non-exact to an exact linear differential equation

Quote:

Originally Posted by

**Ackbeet** Try $\displaystyle u(x,y)=h(x^{m}y^{n}).$

It may be more algebra, calculus, and differential equations than other methods, but it takes out some of the guesswork. What do you get with this?

Well sir, I honestly don't know, but I decided to math dangerously. If I take your advice and multiplied both sides of the original equation by $\displaystyle h(x^{m}y^{n}).$, then I get

$\displaystyle h(x^{m}y^{n})M dx + h(x^{m}y^{n})N dy = 0 $

If the equation is exact, then

$\displaystyle (d/dy) h(x^{m}y^{n})M = (d/dx) h(x^{m}y^{n})N$ which implies

$\displaystyle (d/dy)M / (d/dx)N = h(nx/my)$ ?

Re: transforming from a non-exact to an exact linear differential equation

Hmm. Not quite sure I would go there. This is what I do. I do start out with your initial equation there:

$\displaystyle \frac{\partial}{\partial y}\left[h(x^{m}y^{n})M(x,y)\right]=\frac{\partial}{\partial x}\left[h(x^{m}y^{n})N(x,y)\right].$

But then I use the chain and product rules thus:

$\displaystyle ny^{n-1}h'(x^{m}y^{n})M(x,y)+h(x^{m}y^{n})\frac{\partial M(x,y)}{\partial y}=mx^{m-1}h'(x^{m}y^{n})N(x,y)+h(x^{m}y^{n})\frac{\partial N(x,y)}{\partial x}.$

You plug in what $\displaystyle \frac{\partial M(x,y)}{\partial y}$ and $\displaystyle \frac{\partial N(x,y)}{\partial x}$ are for your DE. Then, you let $\displaystyle u=x^{m}y^{n},$ and try to get a differential equation for $\displaystyle h(u)$ that you can solve. Because you have lots of freedom here, you can often choose $\displaystyle n$ and/or $\displaystyle m$ judiciously to greatly simplify your problem.

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Re: transforming from a non-exact to an exact linear differential equation

As suggested by Danny, it is easy to obtain h(y) = exp(y)