Exact Equations & Substitutions

Why level curves: along a solution $y(x)$, the chain rule gives $\dfrac{d}{dx}F(x, y(x)) = F_x + F_y\,y' = M + N y'$, which is zero exactly when the equation holds. So $F$ is constant along every solution. To build $F$: integrate $M$ in $x$ (adding an unknown $g(y)$), differentiate in $y$, and match against $N$ to determine $g$ — the same construction as finding a potential for a conservative vector field, because it is the same problem.

Example: $y\,dx + 2x\,dy$ has $M_y = 1 \ne 2 = N_x$, but $\mu = y$ turns it into $y^2dx + 2xy\,dy = d(xy^2)$ — exact. The linear-equation integrating factor from first-order theory is a special case of this idea. Where the formula comes from: with $\mu = \mu(x)$, exactness of $\mu M\,dx + \mu N\,dy$ reads $\mu M_y = \mu' N + \mu N_x$, i.e. $\dfrac{\mu'}{\mu} = \dfrac{M_y - N_x}{N}$. The left side depends on $x$ only, so this can be solved for $\mu(x)$ exactly when the right side does too.

Both are one-line changes of viewpoint that convert something unsolvable-looking into a solved chapter; recognizing the pattern is the entire skill. The mechanism in the homogeneous case: $y = vx$ gives $y' = v + xv'$ by the product rule, while the right side becomes $f(v)$, so $xv' = f(v) - v$ — the variables separate.

Step 2 always produces an equation for $g'$ with no $x$ in it, and that is exactly where exactness is used: $N - \partial_y\!\int M\,dx$ has $x$-derivative $N_x - M_y = 0$, so it depends on $y$ alone. If an $x$ survives in $g'(y)$, the equation was not exact or an integration slipped. The solution through $(x_0, y_0)$ is the level curve $F = F(x_0, y_0)$; read as a function $y(x)$, it continues until the curve turns vertical, which happens where $N = 0$.

Step 3 — match: $F_y = x + g'(y)$ must equal $N = x + 2y$, so $g' = 2y$, $g = y^2$ and $F = x^2 + xy + y^2$.

Step 4 — the data: $F(1, 1) = 3$, so the solution curve is $x^2 + xy + y^2 = 3$, an ellipse.

Step 5 — as a function: solving the quadratic in $y$ gives $y = \dfrac{-x \pm\sqrt{12 - 3x^2}}{2}$, and the $+$ sign is the one with $y(1) = \dfrac{-1 + 3}{2} = 1$.

Step 6 — where it lives: the tangent turns vertical where $N = x + 2y = 0$, which on the ellipse happens at $(2, -1)$ and $(-2, 1)$. The solution exists on $-2 < x < 2$; the lower half of the ellipse is a different solution, the one through $(1, -2)$.

Step 3 — match: $F_y = xe^{xy} + g'(y)$ must equal $xe^{xy} - 2y$, so $g' = -2y$ and $g = -y^2$.

Step 4 — $F = e^{xy} + x^2 - y^2$, and $F(0, 1) = 1 + 0 - 1 = 0$, so the solution is the curve $e^{xy} + x^2 - y^2 = 0$.

Step 5 — verify by implicit differentiation: $e^{xy}(y + xy') + 2x - 2yy' = 0$ rearranges to $(ye^{xy} + 2x) + (xe^{xy} - 2y)y' = 0$ ✓. No formula for $y$ in elementary functions exists here; the implicit curve is the answer.

Each substitution turns the troublesome combination into a single new unknown. In the homogeneous case the right side depends on $x$ and $y$ only through $v = \tfrac yx$, and $y' = v + xv'$ leaves $x$ appearing only as a factor, so the variables separate. In the Bernoulli case, $y^{-n}y'$ is, up to the constant $1 - n$, the derivative of $y^{1-n}$, so dividing by $y^n$ makes $v = y^{1-n}$ appear linearly. In the third case the right side is constant along the lines $ax + by + c = u$, so tracking $u$ alone is enough. After solving, substitute back and check against the original equation.

Step 3 — separate: $e^{v}\,dv = \dfrac{dx}{x}$, so $e^{v} = \ln x + C$.

Step 4 — the data: $v(1) = 0$ gives $1 = 0 + C$. Then $e^{v} = 1 + \ln x$, $v = \ln(1 + \ln x)$, and $y = x\ln(1 + \ln x)$.

Step 5 — interval: the logarithm needs $1 + \ln x > 0$, so the solution lives on $x > e^{-1}$.

Step 6 — verify: $y' = \ln(1 + \ln x) + \dfrac{1}{1 + \ln x}$, and $\dfrac yx + e^{-y/x} = \ln(1 + \ln x) + \dfrac{1}{1 + \ln x}$ ✓.

Step 3 — integrating factor $\mu = e^{-\ln x} = \dfrac1x$: $\left(\dfrac vx\right)' = -1$, so $\dfrac vx = -x + C$ and $v = Cx - x^2$.

Step 4 — the data: $v(1) = 1$ gives $C = 2$, so $v = 2x - x^2$.

Step 5 — back to $y$: $y = \dfrac{1}{2x - x^2} = \dfrac{1}{x(2 - x)}$.

Step 6 — interval: $v$ vanishes at $x = 0$ and $x = 2$, so the solution lives on $0 < x < 2$ and blows up at both ends, although the equation is perfectly smooth at $x = 2$.

Step 7 — verify: $y' = \dfrac{2x - 2}{(2x - x^2)^2}$, and $xy^2 - \dfrac yx = \dfrac{x - (2 - x)}{(2x - x^2)^2} = \dfrac{2x - 2}{(2x - x^2)^2}$ ✓.

Step 3 — $\displaystyle\int\frac{du}{1 + u^2} = \int dx$ gives $\arctan u = x + C$, and $u(0) = 0$ gives $C = 0$.

Step 4 — so $u = \tan x$ and $y = \tan x - x$, on $|x| < \tfrac\pi2$.

Step 5 — verify: $y' = \sec^2 x - 1 = \tan^2 x = (x + y)^2$ ✓.

Conversely, on a rectangle define $F(x,y) = \int_{x_0}^x M(t, y)\,dt + \int_{y_0}^y N(x_0, s)\,ds$; then $F_x = M$ directly, and differentiating under the integral, $F_y = \int_{x_0}^x M_y\,dt + N(x_0, y) = \int_{x_0}^x N_x\,dt + N(x_0,y) = N(x, y)$, using $M_y = N_x$ in the middle step. $\blacksquare$

Step 1 — test: $M_y = 3x + 2y$, $N_x = 2x + y$ — not exact.

Step 2 — try a factor in $x$: $\dfrac{M_y - N_x}{N} = \dfrac{x + y}{x(x + y)} = \dfrac1x$, a function of $x$ alone, so $\mu = e^{\int dx/x} = x$.

Step 3 — multiply: $(3x^2y + xy^2)\,dx + (x^3 + x^2y)\,dy = 0$; now $M_y = 3x^2 + 2xy = N_x$ ✓.

Step 4 — build $F$: integrate $M$ in $x$: $F = x^3y + \tfrac12x^2y^2 + g(y)$; match $F_y = x^3 + x^2y + g'(y) = N$, so $g' = 0$. Solutions: $x^3y + \tfrac12x^2y^2 = C$.

Step 5 — implicit differentiation of the answer reproduces the original equation (after dividing by $x$) ✓.

Step 1 — every term has total degree $2$ over degree $2$: homogeneous type. Rewrite $y' = \dfrac xy + \dfrac yx$ and substitute $v = \dfrac yx$, so $y' = v + xv'$.

Step 2 — the equation becomes $v + xv' = \dfrac1v + v$, i.e. $xv' = \dfrac1v$: separable.

Step 3 — $v\,dv = \dfrac{dx}{x}$ gives $\tfrac{v^2}{2} = \ln|x| + C$.

Step 4 — back-substitute: $\dfrac{y^2}{2x^2} = \ln|x| + C$; the data $y(1) = 2$ give $2 = 0 + C$. Solution: $y^2 = 2x^2(\ln x + 2)$, i.e. $y = x\sqrt{2\ln x + 4}$ for $x > 0$.

Step 5 — check at $x = 1$: $y = \sqrt4 = 2$ ✓.

Step 1 — test: $\partial_y(2xy + 3) = 2x$ and $\partial_x(x^2 + 4y) = 2x$: exact.

Step 2 — integrate $M$ in $x$: $F = x^2y + 3x + g(y)$.

Step 3 — match: $F_y = x^2 + g'(y) \stackrel{!}{=} x^2 + 4y$, so $g'(y) = 4y$, $g = 2y^2$.

Step 4 — solutions: $x^2y + 3x + 2y^2 = C$, an implicit family; a point condition picks the curve.

Step 5 — verify by implicit differentiation: $2xy + x^2y' + 3 + 4yy' = 0$ rearranges to the original equation ✓.

Step 1 — divide by $y^2$: $y^{-2}y' + y^{-1} = 1$.

Step 2 — substitute $v = y^{-1}$, $v' = -y^{-2}y'$: $-v' + v = 1$, i.e. $v' - v = -1$.

Step 3 — integrating factor $e^{-x}$: $(ve^{-x})' = -e^{-x}$, so $ve^{-x} = e^{-x} + C$ and $v = 1 + Ce^{x}$.

Step 4 — data: $v(0) = 2$ gives $C = 1$: $y = \dfrac{1}{1 + e^x}$.

Step 5 — verify: $y' = \dfrac{-e^x}{(1+e^x)^2}$ and $y^2 - y = \dfrac{1 - (1 + e^x)}{(1+e^x)^2} = \dfrac{-e^x}{(1+e^x)^2}$ ✓.