Exact Equations

Note that this equation is written in a slightly different form but we can rewrite it as follows:

\( (y \cos x + 3x^2 e^y) + (\sin x + x^3 e^y)y^{'}\)

\( (y \cos x + 3x^2 e^y) + (\sin x + x^3 e^y)\cfrac{dy}{dx}\)

\( (y \cos x + 3x^2 e^y)dx + (\sin x + x^3 e^y)dy\)

We identify \(M(x,y)\) and \(N(x,y)\):

\(M(x,y) = y \cos x + 3x^2 e^y\)

\(N(x,y) = \sin x + x^3 e^y\)

Now check \( \cfrac{\partial M}{\partial y} = \cfrac{\partial N}{\partial x} \):

\(M_y = \cos x + 3x^2 e^y\)

\(N_x = \cos x + 3x^2 e^y\)

Since, \( \cfrac{\partial M}{\partial y} = \cfrac{\partial N}{\partial x} \), the expression is an exact differential.

The equation is in the already in the form \( M(x,y)dx + N(x,y)dy = 0\). \(M(x,y) = 3x^2-2xy+2\). We integrate wrt to \(x\).

\( \int (3x^2-2xy+2)dx = x^3 -x^2y + 2x + h(y)\)

Now take the derivative wrt \(y\) and set to equal to \(N(x,y)\).

\( \cfrac{\partial}{\partial y} (x^3 - x^2y + 2x + h(y)) = 0 - x^2 + 0 + h^{'}(y)= -x^2 + h^{'}(y) \)

\( -x^2 + h^{'}(y) = 6y^2-x^2+3 \)

We can solve for \( h^{'}(y) \) and then intagrate wrt \(y\).

\( h^{'}(y) = 6y^2 + 3 \)

\( h (y) = \int (6y^2 + 3) dy \)

\( h (y) = 2y^3 + 3y \)

Substituting \(h(y)\) back into the first step yields:

\( x^3 -x^2y + 2x + 2y^3 + 3y = c\)

We can confirm the solution:

\(M(x,y) = \cfrac{\partial f}{\partial x} = \cfrac{\partial}{\partial x} (x^3 -x^2y + 2x + 2y^3 + 3y) = 3x^2 - 2xy + 2\)

\(N(x,y) = \cfrac{\partial f}{\partial y} = \cfrac{\partial}{\partial y} (x^3 -x^2y + 2x + 2y^3 + 3y) -x^2 + 6y^2 + 3\)