Exact Equations — Question 1

Question 1 – Solution

Strategy. Recover a potential from both coefficients, then examine its derivative with respect to $y$ on the selected level.

Step 1: Test and integrate. Here $M_y=2x=N_x$, so the form is exact on $\mathbb R^2$. Integrating $F_x=M$ gives $$F=x^2y+x+h(y),\qquad F_y=x^2+h'(y).$$ Matching $N$ requires $h'(y)=y^2$, so we may choose $F=x^2y+x+y^3/3$. At $(1,1)$, $F=7/3$. Thus $$\boxed{x^2y+x+\frac{y^3}{3}=\frac 73}.$$

Step 2: Verify and find the tangent. Differentiating this identity gives $$2xy+1+(x^2+y^2)y'=0,$$ exactly the original equation for a graph. At $(1,1)$ this yields $y'=-3/2$, hence $\boxed{y-1=-\tfrac 32(x-1)}$.

Step 3: Establish a global graph. For fixed $x$, the function $u\mapsto x^2u+u^3/3$ is strictly increasing and ranges over all real numbers. This remains true at $x=0$, where the cubic alone is strictly increasing. Therefore the equation $$x^2u+\frac{u^3}{3}=\frac 73-x$$ has exactly one real root for every real $x$.

The derivative $F_y=x^2+y^2$ can vanish only at $(0,0)$, but that point has $F=0$, not $7/3$. It is consequently nonzero everywhere on the selected level. The implicit function theorem gives a differentiable local graph at each point; the uniqueness of each root makes those graphs agree. They form one differentiable solution on all of $\mathbb R$. The isolated degeneracy on a different level does not obstruct this IVP.