Exact Equations: Question 2-3.10 & Solution

Question 10

For the exact form $\omega=(2xy+e^x)\,dx+(x^2+2y)\,dy,$ compare two routes from $A=(0,0)$ to $B=(1,1)$ . Route 1 goes horizontally to $(1,0)$ and then vertically to $B$ ; Route 2 goes vertically to $(0,1)$ and then horizontally to $B$ .

For an axis-aligned segment, define its accumulated value as the integral of the coefficient of $dx$ along a horizontal segment, or of the coefficient of $dy$ along a vertical segment, with the stated orientation.

Tasks

Construct a potential $F$ normalized by $F(0,0)=0$ .
Compute both route totals directly as sums of ordinary single-variable integrals.
Explain their equality using the chain rule and find the total around the closed route that follows Route 1 forward and Route 2 backward.
For the differential equation $\omega=0$ , decide whether one solution curve could pass through both $A$ and $B$ . Distinguish this question from the equality of the two route totals.

Original worksheet page 1: question and worked solution for 2-3-010

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Question 10 – Solution

Strategy. A potential measures accumulated change along a route; a solution of the equation $\omega=0$ must instead keep that potential constant.

Step 1: Construct a normalized potential. The cross partials are $2x$ . Integrating $M$ gives $F=x^2y+e^x+h(y)$ ; matching $N$ requires $h'=2y$ . Normalization yields $\boxed{F=x^2y+e^x+y^2-1},\qquad F(B)=e+1.$

Step 2: Compute the routes directly. On Route 1, the horizontal and vertical contributions are $\int_0^1 e^x\,dx=e-1,\qquad \int_0^1(1+2y)\,dy=2,$ so its total is $e+1$ . On Route 2 they are $\int_0^1 2y\,dy=1,\qquad \int_0^1(2x+e^x)\,dx=e,$ again totaling $\boxed{e+1}$ .

Step 3: Explain equality and the closed route. Along a differentiable parametrized segment, $dF/dt=F_x\,x'(t)+F_y\,y'(t)=Mx'+Ny'$ . Integration gives the endpoint difference of $F$ ; these differences telescope across joined segments. Both routes therefore give $F(B)-F(A)=e+1$ . Reversing a route negates its integrals, so the closed-route total is $\boxed{0}$ .

Step 4: Distinguish routes from solutions. A solution curve of $\omega=0$ has $dF/dt=0$ and stays on one level. Since $F(A)=0$ and $F(B)=e+1\ne 0$ , no such curve can pass through both points. Equal route totals express endpoint dependence; they do not make either route a solution of the differential equation.

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 2-3-010

Exact Equations — Question 10

Question 10

Question 10 – Solution