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:
$\displaystyle M(x,y)~dx + N(x,y)~dy = 0$
here $\displaystyle M$ and $\displaystyle N$ are functions of $\displaystyle x$ and $\displaystyle y$.
to be exact, we required that $\displaystyle M_y = N_x$, that is, $\displaystyle \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}$
now i'm off to bed
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
$\displaystyle (6xy^3+ cos(y)+ 2kx^2y^3- xsin(y))dx+ ydy= 0$.
An equation of the form M(x,y)dx+ N(x,y)dy= 0 is "exact" if and only if $\displaystyle M_y= N_x$
Here $\displaystyle M= 6xy^3+ cos(y)+ 2kx^2y^3- xsin(y)$ so [tex]M_y= 18xy^2- sin(y)+ 6kx^2y^2- xcos(y)[tex] while $\displaystyle N= y$ so $\displaystyle N_x= 0$. There is NO value of k that will make that exact.
Did you mean $\displaystyle 6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y) dy/dx=0$?
In that case $\displaystyle M= 6xy^3+ cos(y)$ so $\displaystyle M_y= 18xy^2- sin(y)$ and $\displaystyle N= 2kx^2y^2- x sin(y)+ y$ so $\displaystyle N_x= 4kxy^2- sin(y)$. 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 $\displaystyle dF= 6xy^3+cos(y)+(2kx^2y^2-xsin(y)+y) dy/dx=0$. We must have $\displaystyle F_x= 6xy^3+ cos(y)$. Integrating that with respect to x while holding y constant, $\displaystyle F(x,y)= 3x^2y^3+ xcos(y)+ p(y)$. 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, $\displaystyle F_y= 9x^2y^2- xsin(y)+ p'(y)$ and that must be equal to $\displaystyle 2kx^2y^2- xsin(y)+ y$ which, since k= 9/2 says $\displaystyle 9x^2y^2- xsin(y)+ p'(y)= 9x^2y^2- xsin(y)+ y$ so that everything involving x cancels (as it must) and we have p'(y)= y. From that $\displaystyle p(y)= \frac{1}{2}y^2+ C$ where C now really is a constant.
Then $\displaystyle F(x,y)= x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2+ C$.
The original equation says that dF= 0 which means F is equal to a constant: $\displaystyle F(x,y)= x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2+ C= C'$ and, combining the fractions, $\displaystyle x^2y^3+ xcos(y)+ p(y)= x^2y^3+ xcos(y)+ \frac{1}{2}y^2= C$.