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

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 only!

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

Quote:

Originally Posted by

**Ackbeet** Try

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 , then I get

If the equation is exact, then

which implies

?

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:

But then I use the chain and product rules thus:

You plug in what and are for your DE. Then, you let and try to get a differential equation for that you can solve. Because you have lots of freedom here, you can often choose and/or 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)