Checking my solution to an exact differential equation problem

[mr_coffee]

Hello everyone, its exam time so I'm just reviewing all the different techniques for Diff EQ and i hit an exact diff EQ problem and i was just wondering if someone can check to see if i did it right. HEre it is:
Directions are find a relationships between x and y of the form f(x,y) = c.
http://suprfile.com/src/1/27rafy/lastscan.jpg

Thanks!

 

[quasar987]

Yes, except for the line $\Psi(x,y) = \int Mdx$ which should read $\Psi(x,y) = \int Mdx + h(y)$, it's good. But in an exam, you might want to explain what and why you do these things more.

 

[mr_coffee]

Awesome thanks quasar

 

[HallsofIvy]

On the contrary, a lot of what you have written is non-sense!

You are correct that, with $\Phi_x$= 3x²-2xy+ 2, you must have $\Phi$= x³- x²y+ 2x+ h(y). The derivative of that, with respect to y is -x²+ h'(y)
and, of course, that must equal N which is 6y²- x²+ 3, not 6y²- x³+ 3!

You should have seen that you copied that incorrectly: you have
-x²+ h'(y)= 6y²- x³+ 3 and then h'(y)= 6y²- x³+ x²+ 3 which is impossible: h is a function of y only!

Writing it correctly, -x²+ h'(y)= 6y²- x²+ 3, h'(y)= 6y²+ 3. The x² terms cancel which has to happen- that's the whole point of the differential being exact!

With h'(y)= 6y²+ 3, h(y)= 2y³+ 3y and the general solution to the differential equation is
x³- x²y+ 2x+ 2y³+ 3y= Constant.

 

[quasar987]

I didn't noticed h(y) had x terms in it!

mr_coffee, like HallsofIvy said, h(y) has to be a function of y. Why? Because if in fact there exists a function $\psi(x,y)$ such that $\partial \psi / \partial x = M$ and such that $\psi(x,y) = \int Mdx + h$, it must be that h is a function of y only because it it were a function of x too, differentiation of that last equation would give $\partial \psi / \partial x = M + \partial h(x,y) / \partial x \neq M$ which contradicts the first equation.

So verifying that h is a function of y only is always a good way to check if you've made some mistakes along the way.

 

[mr_coffee]

Ahh thanks guys!
Here i redid it, i think its right now:
http://suprfile.com/src/1/2deull/lastscan.jpg

 

[HallsofIvy]

Yes, that is correct!