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

1. ## 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'

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

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

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?

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

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

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

Originally Posted by Ackbeet
Try $u(x,y)=h(x^{m}y^{n}).$

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 $h(x^{m}y^{n}).$, then I get

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

If the equation is exact, then
$(d/dy) h(x^{m}y^{n})M = (d/dx) h(x^{m}y^{n})N$ which implies

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

5. ## 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:

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

But then I use the chain and product rules thus:

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

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

6. ## 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)