# Math Help - Proving implicit equations are exact

1. ## Proving implicit equations are exact

I have been given the equation:
6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y dy/dx=0

I need to
(a) find the value of k that will make the equation exact.

(b) Find an implicit solution when y(0)=1

2. Originally Posted by sazzy3
I have been given the equation:
6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y dy/dx=0

I need to
(a) find the value of k that will make the equation exact.

(b) Find an implicit solution when y(0)=1
ok, it seems you are missing parentheses or something. i am simply too tired to figure out where. here is how to proceed. write your expression in the following way:

$M(x,y)~dx + N(x,y)~dy = 0$

here $M$ and $N$ are functions of $x$ and $y$.

to be exact, we required that $M_y = N_x$, that is, $\frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}$

now i'm off to bed

3. ## continued

i get for f(x,t)= 6xy^3 +cos (y) (when you treat y as a constant)
the result of

6y^3/18xy^2

????? is this right because i cant seem to g(x,t)= 2kxy^2- x sin (y) + y to work anywhere near that with a k value when treating x as a constant.

4. Originally Posted by sazzy3
I have been given the equation:
6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y dy/dx=0

I need to
(a) find the value of k that will make the equation exact.

(b) Find an implicit solution when y(0)=1
There is an extra "(" in your equation. If I drop it I have [tex]6xy^3+ cos(y)+ 2kx^2y^2- xsin(y)+ ydy/dx= 0.
"Multiplying" the entire equation by dx gives
$(6xy^3+ cos(y)+ 2kx^2y^3- xsin(y))dx+ ydy= 0$.

An equation of the form M(x,y)dx+ N(x,y)dy= 0 is "exact" if and only if $M_y= N_x$
Here $M= 6xy^3+ cos(y)+ 2kx^2y^3- xsin(y)$ so [tex]M_y= 18xy^2- sin(y)+ 6kx^2y^2- xcos(y)[tex] while $N= y$ so $N_x= 0$. There is NO value of k that will make that exact.

Did you mean $6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y) dy/dx=0$?
In that case $M= 6xy^3+ cos(y)$ so $M_y= 18xy^2- sin(y)$ and $N= 2kx^2y^2- x sin(y)+ y$ so $N_x= 4kxy^2- sin(y)$. Those will be equal and the equation will be exact if 4k= 18 or if k= 9/2.

In that case, there exist F(x,y) such that $dF= 6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y) dy/dx=0$. We must have $F_x= 6xy^3+ cos(y)$. Integrating that with respect to x while holding y constant, $F(x,y)= 3x^2y^3+ xcos(y)+ p(y)$. The "constant of integration", since we are holding y constant, may be some function of y, p(y).

Now, differentiating that with respect to y, $F_y= 9x^2y^2- xsin(y)+ p'(y)$ and that must be equal to $2kx^2y^2- xsin(y)+ y$ which, since k= 9/2 says $9x^2y^2- xsin(y)+ p'(y)= 9x^2y^2- xsin(y)+ y$ so that everything involving x cancels (as it must) and we have p'(y)= y. From that $p(y)= \frac{1}{2}y^2+ C$ where C now really is a constant.

Then $F(x,y)= x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2+ C$.

The original equation says that dF= 0 which means F is equal to a constant: $F(x,y)= x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2+ C= C'$ and, combining the fractions, $x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2= C$.