Bernoulli & Exact Equations – Detailed Review

Review – Homogeneous Differential Equations

• Criterion for homogeneity
  • Each term in the ODE must have the same total degree in its variables.
  • Works even if the variables are different (e.g., $x$ in one term, $y$ in another) as long as the sum of exponents in each term is identical.
  • Example with a square‐root term: you can factor out $x^2$ or $y^2$ to reveal common degree; homogeneity is preserved.
• Practical test
  • Replace $x \to kx$ and $y \to ky$.
  • If the whole expression scales by $k^m$ for some single $m$, the equation is homogeneous.
• Standard substitution for homogeneous first–order ODEs
  • Set $v = \frac{y}{x} \;\Rightarrow\; y = vx\; ,\; \frac{dy}{dx} = v + x\frac{dv}{dx}$.
• Goal after substitution: convert to a separable ODE and integrate ("antiderivative frenzy").

Bernoulli Equations

• Canonical form
  $\frac{dy}{dx} + P(x)\,y = Q(x)\,y^{n}, \qquad n \ne 0,1$
• Transformation that makes it linear
  • Let $v = y^{1-n} \;\;(\text{or } v = y^{1-n})$.
  • Then $\frac{dv}{dx} = (1-n)\,y^{-n}\,\frac{dy}{dx}$.
  • Substituting converts the Bernoulli equation into
    $\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x)$
    which is linear in $v$.
• Solve the linear equation with the integrating‐factor method
  • Integrating factor
    $\mu(x)=\exp\left(\int (1-n)P(x)\,dx\right)$
  • General solution in $v$, then back‐substitute $y = v^{\,1/(1-n)}$.

Walk-Through Example (n = −1)

1. Starting ODE (after instructor divided by $2xy$):
   $\frac{dy}{dx}-\frac{1}{y}= \frac{1}{x}\,y^{-1}$ (implicitly $n=-1$).
2. Identify parameters
   $P(x)=-\frac{1}{y} \;\;(P \text{ here is actually a constant wrt }x), \qquad Q(x)=\frac{1}{x}, \qquad n=-1.$
3. Bernoulli substitution $v=y^{1-(-1)}=y^{2}$.
4. Derivative link $\frac{dv}{dx}=2y\frac{dy}{dx}$; solve $\frac{dy}{dx}$ from the ODE and replace.
5. Obtain the linear $dv/dx$ equation, find integrating factor $\mu(x)$, integrate once, then substitute back $y=\sqrt{v}\,.$
6. Final implicit solution: typical form $y^{2} = C\,x^{\alpha} + f(x)$ with constants determined by integration.

Secondary Example (n = 4/3)

• Instructor identified $n=\frac{4}{3}$ during board work.
• Substitution: $v = y^{1-\frac{4}{3}} = y^{-\frac{1}{3}}$.
• Derived derivative: $\frac{dv}{dx} = -\frac{1}{3} y^{-\frac{4}{3}}\frac{dy}{dx}$.
• After algebra: a linear ODE in $v$ of the form
  $\frac{dv}{dx}+\Bigl(-\tfrac{4}{3}\cdot\text{(some coefficient)}\Bigr)v = \text{rhs}(x)$.
• Same integrating‐factor steps follow.

Exam & Study Tips Mentioned

• "Antiderivative" work pervades this course — every technique aims to produce an integral you can evaluate.
• When you open any new problem, mentally check the hierarchy:
  1. Is the ODE separable?
  2. Is it homogeneous (use $v=y/x$)?
  3. Does it match Bernoulli form?
  4. Could it be exact?
• Expected test timing: if a Bernoulli or homogeneous problem is allocated $\sim7$ min on Test 1, plan for $\sim5$ min on the final.
• Homework counts 85 % of its stated value; turn it in before in-class review for any credit.

Exact Differential Equations

• Structure
  $M(x,y)\,dx + N(x,y)\,dy = 0.$
• Exactness criterion
  $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \quad \text{on a simply-connected region }R=[a,b]\times[c,d].$
• Consequence: there exists a potential function $F(x,y)$ such that
  $dF = \frac{\partial F}{\partial x}\,dx + \frac{\partial F}{\partial y}\,dy = M\,dx + N\,dy.$
• Therefore $M = \frac{\partial F}{\partial x}, \; N = \frac{\partial F}{\partial y}.$
• Solving algorithm
  1. Verify exactness using the partial-derivative test.
  2. Integrate $M$ w.r.t. $x$ (treat $y$ as constant) to get
     $F(x,y) = \int M\,dx + C_1(y).$
     • Note: the "constant" of integration becomes an unknown function of $y$ because partial integration regards $y$ as a parameter.
  3. Differentiate this preliminary $F$ with respect to $y$: $\partial F/\partial y.$
  4. Set the result equal to the given $N(x,y)$ and solve for $C<em>1'(y)$; integrate to discover $C</em>1(y).$
  5. Write the implicit solution $F(x,y)=C$ (true constant).
• Mini-Example worked on the board
  • Given $M = 3x^2y - x y^3$ and $N = \dots$ (full $N$ wasn’t captured in transcript).
  • Integration w.r.t $x$ produced
    $F(x,y)=3x^{2}y - x y^{3} + C_1(y).$
  • Differentiate w.r.t $y$, equate to $N$, determine $C_1(y)$, then state $F=C.$
  • Key takeaway: $C_1$ is a function of $y$ only, not a pure constant.

Multivariable & Line-Integral Connection

• Exact equations mirror the gradient theorem in calculus III.
• In line integrals, if a vector field $\mathbf{F}=\langle M,N\rangle$ is conservative, then the work integral along any path only depends on end-points — same potential function $F(x,y).$
• The differential-equation test $\partial M/\partial y = \partial N/\partial x$ is identical to the curl-free condition for 2-D vector fields.

Real-World Contexts Mentioned

• Flight trajectories in tightly managed airspace
  • Airlines follow pre-computed "logarithm-like" level paths to guarantee safe gaps; underlying calculations are differential-equation based.
• Water current vs. swimmer problems (classic relative-motion ODEs).
• Economic models — rapid parameter changes often modeled with first-order DEs.

Common Pitfalls & Instructor Comments

• Do not confuse separable, homogeneous, and exact; each has its own fingerprint.
• Fractions can obscure pattern recognition; consider clearing denominators early.
• Double-inverse trigonometric substitutions (e.g., undoing $\arctan$ then applying $\tan$ again) might appear but probably avoided on early tests.
• Practice reduces solution time; otherwise algebraic slips (especially with exponents like $\frac{4}{3}$) can derail the entire problem.

Quick Reference Formulas

• Bernoulli integrating factor:
  $\mu(x)=e^{\int(1-n)P(x)\,dx}$
• Exactness test:
  $\boxed{\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}}$
• Potential function reconstruction:
  $F(x,y)=\int M\,dx + C<em>1(y), \qquad C</em>1'(y)=N-\frac{\partial}{\partial y}\left(\int M\,dx\right).$