Ordinary Differential Equations 5 | Solve First-Order Autonomous Equations

TL;DR

Autonomous first-order ODEs ẋ = V(x) with x(0)=x0 can be solved by dividing by V(x) near x0 when V(x0)≠0.

Briefing Cornell Notes

Briefing

First-order autonomous differential equations admit a general, systematic solution method—provided the function driving the dynamics doesn’t vanish at the initial value. For an equation of the form ẋ = V(x) with a continuous V on ℝ and an initial condition x(0)=x0, the key move is to rewrite the ODE as ẋ/V(x)=1 near x0 (assuming V(x0)≠0). That turns the problem into an integrable relationship: integrating from 0 to t yields an equation involving an antiderivative of 1/V(x). Concretely, if F is any antiderivative of 1/V, then the solution satisfies F(x(t)) = t + constant. Solving for x(t) requires inverting F locally, giving x(t) = F^{-1}(t − C), where the constant C is fixed by enforcing x(0)=x0. This framework matters because it converts a differential equation into an algebraic inversion problem—often the difference between a solvable model and one that stays stuck in formal notation.

The transcript then grounds the method with two worked examples that show how the “antiderivative + inverse + constant” workflow plays out in practice. For ẋ = Λx with Λ>0 and x0≠0, separating variables leads to ∫(1/x)dx = ∫Λ dt. The left side becomes ln|x|, so ln|x(t)| = Λt + C. Exponentiating gives |x(t)| = e^{C}e^{Λt}, and the absolute value splits the solution into sign-consistent branches. Applying x(0)=x0 determines the constant so the final solution collapses to the familiar exponential form x(t)=x0 e^{Λt}. The treatment also notes that the absolute value doesn’t create ambiguity once the initial sign is fixed: the exponential factor stays positive, so the solution either remains always positive or always negative depending on x0.

The second example, ẋ = x^2 with x0≠0, produces a rational solution. Separation yields ∫x^{-2}dx = ∫dt, so −1/x(t) = t + C. Solving for x(t) gives x(t)=−1/(t+C). Enforcing x(0)=x0 determines C = −1/x0, leading to x(t)=x0/(1−x0 t). Unlike the exponential case, this expression reveals a finite-time blow-up: the denominator hits zero at t = 1/x0, so the solution cannot extend past that time.

Overall, the central insight is that autonomous first-order ODEs reduce to integrating 1/V(x) and inverting the resulting antiderivative—turning dynamics into a solvable “separate, integrate, invert, fit the constant” recipe. The examples also highlight how qualitative behavior (growth vs. blow-up) emerges directly from the algebraic form of the solution.

Cornell Notes

Autonomous first-order ODEs of the form ẋ = V(x) with x(0)=x0 can be solved by separating variables when V(x0)≠0. Dividing by V(x) gives ẋ/V(x)=1, which integrates to an equation involving an antiderivative F of 1/V: F(x(t)) = t + constant. The constant is chosen so the initial condition x(0)=x0 holds, and the solution is obtained by locally inverting F: x(t)=F^{-1}(t−C). Two examples show the workflow: ẋ=Λx leads to x(t)=x0 e^{Λt}, while ẋ=x^2 leads to x(t)=x0/(1−x0 t), which blows up when 1−x0 t=0.

Why does the method require V(x0)≠0, and what goes wrong if V(x0)=0?

The derivation divides by V(x), rewriting the ODE as ẋ/V(x)=1. If V(x0)=0, the expression 1/V(x) becomes singular at x0, so the integral used to build F(x) may not exist (or may not be well-behaved) near the initial state. The transcript notes that continuity of V ensures the division makes sense in a neighborhood around x0 only when V(x0) is nonzero; zeros of V far from x0 are not the immediate problem for local solvability.

How does the constant of integration get determined in the general autonomous method?

After integrating, the solution has the form F(x(t)) = t + constant, where F is an antiderivative of 1/V. Writing the constant as −C gives F(x(t)) = t − C, so x(t)=F^{-1}(t−C). The constant C is fixed by enforcing the initial condition: plug in t=0 to get x(0)=x0, which implies F(x0)=−C, hence C=−F(x0).

In the example ẋ = Λx, how does ln|x| appear and how does the absolute value affect the final solution?

Separating variables gives dx/x = Λ dt, so integrating yields ln|x| = Λt + C. Exponentiating produces |x(t)| = e^{C}e^{Λt}. Because e^{Λt}>0 for all t, the sign of x(t) is determined by the initial condition x0. Using x(0)=x0 fixes e^{C} so the final result becomes x(t)=x0 e^{Λt} (the absolute value no longer matters once the sign is set by x0).

For ẋ = x^2, why does the solution take a rational form and what does it imply about blow-up?

Separating variables gives dx/x^2 = dt. Integrating yields −1/x(t) = t + C, so x(t)=−1/(t+C). Applying x(0)=x0 gives C=−1/x0, leading to x(t)=x0/(1−x0 t). The denominator 1−x0 t becomes zero at t=1/x0, so the solution becomes unbounded there and cannot be extended past that finite time.

What is the practical “recipe” for solving an autonomous first-order ODE using this approach?

1) Start with ẋ = V(x) and check V(x0)≠0. 2) Divide to get ẋ/V(x)=1. 3) Integrate: ∫(1/V(x))dx = ∫dt, producing F(x(t)) = t + constant. 4) Use x(0)=x0 to determine the constant. 5) Invert F locally to solve for x(t).

Review Questions

Given ẋ = V(x) with x(0)=x0 and V(x0)≠0, what equation involving an antiderivative F of 1/V(x) must x(t) satisfy?
For ẋ = Λx with Λ>0 and x0≠0, what is the explicit solution and why does the sign of x(t) stay fixed?
For ẋ = x^2 with x0≠0, at what time does the solution blow up, and how is that time read from the formula x(t)=x0/(1−x0 t)?

Key Points

1
Autonomous first-order ODEs ẋ = V(x) with x(0)=x0 can be solved by dividing by V(x) near x0 when V(x0)≠0.
2
Integrating ẋ/V(x)=1 leads to an antiderivative relation F(x(t)) = t + constant, where F′(x)=1/V(x).
3
The integration constant is fixed by enforcing x(0)=x0, typically giving C=−F(x0) in the form x(t)=F^{-1}(t−C).
4
For ẋ = Λx (Λ>0), separation yields ln|x| = Λt + C and the solution simplifies to x(t)=x0 e^{Λt}.
5
For ẋ = x^2, separation yields −1/x(t)=t+C and the solution becomes x(t)=x0/(1−x0 t).
6
Rational solutions like x0/(1−x0 t) can blow up at finite time when the denominator reaches zero.
7
The method is fundamentally local: it relies on being able to invert F near the initial value.

Highlights

The general solution strategy reduces solving ẋ=V(x) to integrating 1/V(x), then inverting an antiderivative F locally.

In the linear case ẋ=Λx, the absolute value from ln|x| disappears once x(0)=x0 fixes the sign, yielding x(t)=x0 e^{Λt}.

In the quadratic case ẋ=x^2, the solution x(t)=x0/(1−x0 t) makes finite-time blow-up explicit at t=1/x0.

Topics

Initial Value Problems
Autonomous ODEs
Separation of Variables
Antiderivatives and Inverses
Exponential and Blow-Up Solutions