Exact Equations & Substitutions

Study Sheet

Exact Equations & Substitutions

Exactness tests, integrating factors, Bernoulli and homogeneous types

Exactness and the Great Substitutions

Tip
Exact equations: one function behind two coefficients

The form M(x,y) dx+N(x,y) dy=0M(x,y)\,dx + N(x,y)\,dy = 0 is EXACT when some FF has Fx=MF_x = M and Fy=NF_y = N — then solutions are simply the level curves F(x,y)=CF(x, y) = C. The test is Clairaut in disguise: exactness holds iff ∂yM=∂xN\partial_y M = \partial_x N (both equal the mixed partial FxyF_{xy}).

To build FF: integrate MM in xx (adding an unknown g(y)g(y)), differentiate in yy, and match against NN to determine gg — the same construction as finding a potential for a conservative vector field, because it IS the same problem.

Tip
When the test fails: integrating factors

A non-exact equation can often be repaired by multiplying through by μ\mu. Try single-variable factors first: if My−NxN\dfrac{M_y - N_x}{N} depends on xx alone, μ=e∫My−NxNdx\mu = e^{\int\frac{M_y - N_x}{N}dx} works; if Nx−MyM\dfrac{N_x - M_y}{M} depends on yy alone, use the analogous μ(y)\mu(y).

Example: y dx+2x dyy\,dx + 2x\,dy has My=1≠2=NxM_y = 1 \ne 2 = N_x, but μ=y\mu = y turns it into y2dx+2xy dy=d(xy2)y^2dx + 2xy\,dy = d(xy^2) — exact. The linear-equation integrating factor from first-order theory is a special case of this idea.

Tip
Two substitutions that unlock whole families

HOMOGENEOUS type y′=f ⁣(yx)y' = f\!\left(\tfrac yx\right): substitute v=yxv = \tfrac yx (so y′=v+xv′y' = v + xv') and the equation SEPARATES in vv and xx. BERNOULLI y′+p(x)y=q(x)yny' + p(x)y = q(x)y^n: divide by yny^n and substitute v=y1−nv = y^{1-n} — the equation becomes LINEAR in vv, solvable by integrating factor.

Both are one-line changes of viewpoint that convert something unsolvable-looking into a solved chapter; recognizing the pattern is the entire skill.

Side note
Significance: thermodynamics runs on exactness

In physics, "is M dx+N dyM\,dx + N\,dy exact?" becomes "is this quantity a state function?" Energy and entropy are exact differentials (path-independent); heat and work are NOT — and integrating factors are precisely how temperature turns the inexact heat differential into exact entropy: dS=δQTdS = \tfrac{\delta Q}{T}. A chapter of calculus explaining a law of nature.

Try it
Try it: test, then decide

Is (y2+1) dx+2xy dy=0(y^2 + 1)\,dx + 2xy\,dy = 0 exact? Work: ∂y(y2+1)=2y=∂x(2xy)\partial_y(y^2 + 1) = 2y = \partial_x(2xy) ✓. Solve: F=xy2+xF = xy^2 + x; solutions x(y2+1)=Cx(y^2 + 1) = C. Verify by implicit differentiation: (y2+1)+2xyy′=0(y^2+1) + 2xyy' = 0 ✓ — matches the original.

Proofs & Why It Matters

Tip
Proof: the exactness test

If FF exists with Fx=MF_x = M, Fy=NF_y = N and FF is twice continuously differentiable, then My=Fxy=Fyx=NxM_y = F_{xy} = F_{yx} = N_x by Clairaut's theorem — necessity.

Conversely, on a rectangle define F(x,y)=∫x0xM(t,y) dt+∫y0yN(x0,s) dsF(x,y) = \int_{x_0}^x M(t, y)\,dt + \int_{y_0}^y N(x_0, s)\,ds; then Fx=MF_x = M directly, and differentiating under the integral, Fy=∫x0xMy dt+N(x0,y)=∫x0xNx dt+N(x0,y)=N(x,y)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 My=NxM_y = N_x in the middle step. ■\blacksquare

Tip
Proof: the Bernoulli substitution linearizes

From v=y1−nv = y^{1-n}: v′=(1−n)y−ny′v' = (1-n)y^{-n}y'. Multiply the equation y′+py=qyny' + py = qy^n by (1−n)y−n(1-n)y^{-n}: (1−n)y−ny′+(1−n)py1−n=(1−n)q(1-n)y^{-n}y' + (1-n)py^{1-n} = (1-n)q, which is exactly v′+(1−n)p v=(1−n)qv' + (1-n)p\,v = (1-n)q — linear in vv with known coefficient functions. Every Bernoulli equation is therefore one substitution away from the integrating-factor method. ■\blacksquare

Going Deeper: Worked Problems

Example
Worked: repairing a non-exact equation

Solve (3xy+y2) dx+(x2+xy) dy=0(3xy + y^2)\,dx + (x^2 + xy)\,dy = 0.

Step 1 — test: My=3x+2yM_y = 3x + 2y, Nx=2x+yN_x = 2x + y — not exact.

Step 2 — try a factor in xx: My−NxN=x+yx(x+y)=1x\dfrac{M_y - N_x}{N} = \dfrac{x + y}{x(x + y)} = \dfrac1x, a function of xx alone, so μ=e∫dx/x=x\mu = e^{\int dx/x} = x.

Step 3 — multiply: (3x2y+xy2) dx+(x3+x2y) dy=0(3x^2y + xy^2)\,dx + (x^3 + x^2y)\,dy = 0; now My=3x2+2xy=NxM_y = 3x^2 + 2xy = N_x ✓.

Step 4 — build FF: integrate MM in xx: F=x3y+12x2y2+g(y)F = x^3y + \tfrac12x^2y^2 + g(y); match Fy=x3+x2y+g′(y)=NF_y = x^3 + x^2y + g'(y) = N, so g′=0g' = 0. Solutions: x3y+12x2y2=Cx^3y + \tfrac12x^2y^2 = C.

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

Example
Worked: a homogeneous-type equation to the end

Solve y′=x2+y2xyy' = \dfrac{x^2 + y^2}{xy} with y(1)=2y(1) = 2.

Step 1 — every term has total degree 22 over degree 22: homogeneous type. Rewrite y′=xy+yxy' = \dfrac xy + \dfrac yx and substitute v=yxv = \dfrac yx, so y′=v+xv′y' = v + xv'.

Step 2 — the equation becomes v+xv′=1v+vv + xv' = \dfrac1v + v, i.e. xv′=1vxv' = \dfrac1v: separable.

Step 3 — v dv=dxxv\,dv = \dfrac{dx}{x} gives v22=ln⁡∣x∣+C\tfrac{v^2}{2} = \ln|x| + C.

Step 4 — back-substitute: y22x2=ln⁡∣x∣+C\dfrac{y^2}{2x^2} = \ln|x| + C; the data y(1)=2y(1) = 2 give 2=0+C2 = 0 + C. Solution: y2=2x2(ln⁡x+2)y^2 = 2x^2(\ln x + 2), i.e. y=x2ln⁡x+4y = x\sqrt{2\ln x + 4} for x>0x > 0.

Step 5 — check at x=1x = 1: y=4=2y = \sqrt4 = 2 ✓.

Worked, Every Step

Example
Worked: an exact equation start to finish

Solve (2xy+3) dx+(x2+4y) dy=0(2xy + 3)\,dx + (x^2 + 4y)\,dy = 0.

Step 1 — test: ∂y(2xy+3)=2x\partial_y(2xy + 3) = 2x and ∂x(x2+4y)=2x\partial_x(x^2 + 4y) = 2x: exact.

Step 2 — integrate MM in xx: F=x2y+3x+g(y)F = x^2y + 3x + g(y).

Step 3 — match: Fy=x2+g′(y)=!x2+4yF_y = x^2 + g'(y) \stackrel{!}{=} x^2 + 4y, so g′(y)=4yg'(y) = 4y, g=2y2g = 2y^2.

Step 4 — solutions: x2y+3x+2y2=Cx^2y + 3x + 2y^2 = C, an implicit family; a point condition picks the curve.

Step 5 — verify by implicit differentiation: 2xy+x2y′+3+4yy′=02xy + x^2y' + 3 + 4yy' = 0 rearranges to the original equation ✓.

Example
Worked: a Bernoulli equation via v = 1/y

Solve y′+y=y2y' + y = y^2, y(0)=12y(0) = \tfrac12.

Step 1 — divide by y2y^2: y−2y′+y−1=1y^{-2}y' + y^{-1} = 1.

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

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

Step 4 — data: v(0)=2v(0) = 2 gives C=1C = 1: y=11+exy = \dfrac{1}{1 + e^x}.

Step 5 — verify: y′=−ex(1+ex)2y' = \dfrac{-e^x}{(1+e^x)^2} and y2−y=1−(1+ex)(1+ex)2=−ex(1+ex)2y^2 - y = \dfrac{1 - (1 + e^x)}{(1+e^x)^2} = \dfrac{-e^x}{(1+e^x)^2} ✓.