Please check my work (exact equation, then i solved)

[darryw]

Homework Statement

(y-x^3) + (x+y^3)y' = 0

equation is convenienty already in the required form, that is M_(x,y) + N_x(x,y)dy/dx = 0

so..

M_y = 1 = N_x therefore equation is exact. Therefore I now solve...

I am solving for a function, f(x,y) whose partial derivative with respect to y = M
and whose partial with respect to x = N.
Is this correct?

assuming correct so far, my next step is to integrate the M term, with respect to x

integ M_x (y-x^3) = -(1/4)x^4 + h(y)

i have a questions about this step though:

if y's are treated as constants when integrating, wouldn't the y in this equation become yx?

so that integ_x (y-x^3) = yx -(1/4)x^4 + h(y)
??

assuming yx is not in the integrated result, my next step is to take derivative of the just integrated M term, now with respect to y:

derivative of just-integ M_y [-(1/4)x^4 + h(y) ] = h'(y)

so i know that h'(y) = N term which is:

h'(y) = (x+y^3)

But i think I am going to stop here because it looks sort of weird and i want to make sure its right so far.
thanks for any help

Homework Equations

The Attempt at a Solution

 

[Char. Limit]

First, yes, the y in your equation becomes yx. This is true.

Second, now that you know this, keep going! You almost have it, I can tell.

 

[darryw]

ok so slight rework to fix the integration error...

integ M_x (y-x^3) = yx -(1/4)x^4 + h(y)

derivative of just-integ M_y [yx -(1/4)x^4 + h(y) ] = x + h'(y)

x + h'(y) = (x+y^3) (the N term)

h'(y) = y^3

h(y) = (1/4)y^4

correct so far? thank you

 

[darryw]

so, final answer: f(x,y) = x + (1/4)y^4

??

 

[Char. Limit]

\int M_x dx = \int y-x^3 dx = yx - \frac{1}{4}x^4 + h(y)

h(y) = \frac{1}{4}y^4

so, final answer: f(x,y) = x + (1/4)y^4

??

Well, I've cleaned up your answers for latex format, and it looks like your final answer is missing a term. Just look at the first two of the quotes by you and the answer should be right there.

 

[darryw]

for some reason I am confused as to where final answer comes from:
Is it simply N term + whatever I worked out h(y) to be?

If so, answer should be f(x,y) = x + y^3 + (1/4)y^4.

Is this correct and also is answer: N-term + h(y)?
thanks

 

[Mark44]

for some reason I am confused as to where final answer comes from:
Is it simply N term + whatever I worked out h(y) to be?

If so, answer should be f(x,y) = x + y^3 + (1/4)y^4.

Is this correct and also is answer: N-term + h(y)?
thanks

The answer should be f(x, y) = C, where f(x, y) is the function you found by integrating M and N. I haven't checked this work, but the idea is exactly the same as in the other thread you posted.

 

[Char. Limit]

for some reason I am confused as to where final answer comes from:
Is it simply N term + whatever I worked out h(y) to be?

If so, answer should be f(x,y) = x + y^3 + (1/4)y^4.

Is this correct and also is answer: N-term + h(y)?
thanks

Not quite.

The answer is \int \psi_x dx = \psi + h(y)

where psi_x is the partial derivative of psi, which means f(x,y).

You found psi, and you found h(y). Those were the first two items I quoted in my last post.

 

[Mark44]

The answer is \int \psi_x dx = \psi + h(y)
where psi_x is the partial derivative of psi, which means f(x,y).

Since psi means f(x,y), then why not use f(x,y) instead of introducing another symbol that muddies the waters and cannot be typed on standard keyboards?

 

[Char. Limit]

Because I saw psi used as the symbol for f(x,y), and assumed it was common.

Also, how do you write the partial derivative of f(x,y) with respect to x?

Is it {f(x,y)}_x?

 

[Mark44]

The simplest ways use subscripts - f_x or f_x(x, y).

Or you can use this notation
\frac{\partial f}{\partial x}

 

[Char. Limit]

Ah.

Just to clear things up, let me resay what I said before, but in more standard notation:

\int f_x dx = f(x,y) + h(y)

Thanks for clearing that up, Mark. I'll try to be less confusing from now on.

 

[Mark44]

Not a problem.