Exact Equations — Question 3

Question 3 – Solution

Strategy. Exactness requires an identity on a region, so compare the variable-dependent terms before constructing a potential.

Step 1: Determine the parameters. The cross partials are $$M_y=6x^2y+2bx+e^y,\qquad N_x=6ax^2y+10x+e^y.$$ Their difference is $6(1-a)x^2y+(2b-10)x$. For this to vanish for all $(x,y)$, first put $y=0$ and $x=1$, obtaining $b=5$. Then put $x=y=1$, obtaining $a=1$. Conversely, those choices make the difference identically zero. Hence $$\boxed{a=1,\qquad b=5}.$$

Step 2: Construct and verify the potential. With these values, integrating $M$ in $x$ yields $$F=x^3y^2+5x^2y+xe^y+h(y).$$ Matching $F_y$ with $N$ gives $h'(y)=4y^3$. Thus choose $$F=x^3y^2+5x^2y+xe^y+y^4.$$ Direct differentiation gives $F_x=3x^2y^2+10xy+e^y=M$ and $F_y=2x^3y+5x^2+xe^y+4y^3=N$.

Step 3: Apply the data and find the tangent. Since $F(0,1)=1$, the selected level is $\boxed{F(x,y)=1}$. At that point, $N=4\ne 0$, so it determines a local solution graph. Its slope is $-M/N=-e/4$, giving $$\boxed{y-1=-\frac e4x}.$$

Step 4: Explain the failed one-point test. At $(0,0)$, both cross partials equal $1$ for every $a,b$: all parameter-dependent terms vanish there. This agreement is necessary at that point but says nothing about agreement on a neighborhood. Exactness cannot be certified from that single sample.