Find the value of $k$ so that the given differential form is exact?

Question

Find the value of $k$ so that the given differential form is exact.

$$(y^3 + k xy^4 - 2x) dx + (3xy^2 + 20 x^2y^3) dy$$

Answer 1

You want a function $f(x,y)$ such that its differential is the given form (this means to be an exact form), i.e: $$ d f(x,y)=\frac{\partial}{\partial x}f(x,y)dx +\frac{\partial}{\partial y}f(x,y)dy= (y^3+kxy^4-2x)dx +(3xy^2+20x^2y^3)dy $$

In this case it is not difficult to find this function by simple inspection. Note that it must be the sum of three monomial of the form $ ax^ny^m$ and for the first monomial we want: $$ \frac{\partial}{\partial x}ax^ny^m=y^3 \quad \mbox{and} \quad \frac{\partial}{\partial y}ax^ny^m=3xy^2 $$ so the monomial must be $xy^3$ ( the integral with respect to $x$ of $y^3$ that is also the integral with respect to $y$ of $3xy^2$) . In the same way you can see that the third monomial must be $-x^2$.

For the second monomial we have: $$ \frac{\partial}{\partial x}ax^ny^m=kxy^4 \quad \mbox{and} \quad \frac{\partial}{\partial y}ax^ny^m=20x^2y^3 $$

so, integrating the derivative with respect to $y$ we find the monomial $5x^2y^4$ and this means that $k=10$.

So the searched function is $f(xy)=xy^3+5x^2y^4-x^2$

Answer 2

A differential form on Euclidean space is exact if and only if its closed. It's pretty easy to solve the equation

$$\mathrm{d}\left( (y^3 + k xy^4 - 2x) \mathrm{d}x + (3xy^2 + 20 x^2y^3) \mathrm{d}y \right) = 0$$

for $k$, if you actually compute the left hand side.

Answer 3

Hint: For $$ \left(y^3+kxy^4-2x\right)\mathrm{d}x+\left(3xy^2+20x^2y^3\right)\mathrm{d}y $$ to be exact, we need $$ \frac{\partial}{\partial y}\left(y^3+kxy^4-2x\right)=3y^2+4kxy^3 $$ to be equal to $$ \frac{\partial}{\partial x}\left(3xy^2+20x^2y^3\right)=3y^2+40xy^3 $$

Answer 4

$$ P dx+Q dy $$

is exact when

$$ \frac{\partial P} {\partial y} = \frac{\partial Q} {\partial x}$$

Carry out partial differentiations and simplify, $ k=10$.