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For continuously differentiable coefficients on a simply connected domain, the exactness condition is $P_y=Q_x$. Locally the same condition suffices without a global topological assumption. An exact first-order ordinary differential equation has a potential $F$ with $F_x=P$ and $F_y=Q$, so $dF/dx=P+Q{y}^{\prime }=0$ along a solution. Thus its implicit solution is $F(x,y)=C$. One local construction, on a rectangle about $(x_0,y_0)$, isThe equality of cross derivatives verifies both required derivatives; on a simply connected domain one may use the corresponding path-independent integral of the exact differential form.

Here $P=4x+3y$, $Q=3x+3y^2$ and $P_y=Q_x=3$. Integrating $P$ with respect to $x$ gives $F=2x^2+3xy+K(y)$. Matching $F_y=Q$ gives $K^{\prime }=3y^2$, so take $K=y^3$. The initial value sets $C=F(1,2)=16$. ThereforeThis is the requested explicit relation between the coordinates. Since $F_y(1,2)=15\ne 0$, the implicit function theorem gives a unique local graph through the initial point, with slope $-10/15=-2/3$.