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First-Order Equations & Initial Values

Useful preparation: Integrals & the Fundamental Theorem

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A differential equation specifies how a quantity changes at each state, rather than giving the quantity directly. An initial value selects a trajectory from a family. Separation reverses a chain rule; an integrating factor reverses a product rule. Choosing a method means recognizing the structure that makes one of those reversals possible.

Initial-value problem
An equation y prime=f(t,y), together with y (t 0)=y 0\displaystyle y\, \left(t\, 0\right) = y\, 0, asks for a differentiable function satisfying both conditions on an interval containing t0.
Separable equation
An equation y prime=g(t)h(y) separates its dependence on time and state. Where h(y) is nonzero, integrating 1h (y)\displaystyle \frac{1}{h}\, \left(y\right) with respect to y can recover the trajectory.
Linear first-order equation
The form y prime+p (t) y=q (t)\displaystyle p\, \left(t\right)\, y = q\, \left(t\right) is linear in y and its derivative. An integrating factor mu(t)=exp⁡\displaystyle \operatorname{exp}(integral p(t) dt) turns the left side into one product derivative.

Conditions before calculation

  • Before dividing by h(y), record all equilibrium solutions h (y)=0\displaystyle h\, \left(y\right) = 0; otherwise valid constant solutions can disappear.
  • Continuous p and q on an interval give a unique solution of a linear initial-value problem throughout that interval. A leading coefficient must be nonzero before dividing into standard form.
  • For nonlinear problems, continuity of f ensures local existence; local Lipschitz continuity in y ensures local uniqueness. Even smooth equations may have solutions that blow up in finite time.

Why the method works

1. Inspect before integrating

Determine whether the equation separates or is linear. Some equations have both structures; others have neither. Identify the independent variable and units before manipulating symbols.

2. Reverse a differentiation rule

For a linear equation choose mu so that mu prime=p mu. The product rule then supplies a directly integrable derivative.

μ(t)=e∫p(t) dt,(μy)′=μq\mu(t)=e^{\int p(t)\,dt},\qquad (\mu y)'=\mu q

3. Use the initial condition and verify

Integrate, solve for the constant, then check the original derivative equation. Report the interval containing the initial time on which the result remains defined.

Worked examples

EXAMPLE 1

Solve y prime=2t y with y (0)=3\displaystyle y\, \left(0\right) = 3.

  1. The initial value is nonzero. Separating on a nonzero branch gives the logarithm of the absolute value.

    dyy=2t dt,ln⁡∣y∣=t2+C\frac{dy}{y}=2t\,dt,\qquad \ln|y|=t^2+C

  2. Exponentiation gives y=K exp⁡\displaystyle y = K\, \operatorname{exp}(t squared); the initial condition fixes K=3\displaystyle K = 3.

    y(t)=3et2y(t)=3e^{t^2}

  3. Differentiate the result and evaluate at zero to verify both requirements.

    y′=6tet2=2ty,y(0)=3y'=6te^{t^2}=2ty,\qquad y(0)=3

Conclusion: This solution is defined for all real t. The separate equilibrium y=0\displaystyle y = 0 belongs to the differential equation but fails this initial condition.

EXAMPLE 2

Solve y prime+2 y=4\displaystyle 2\, y = 4 with y (0)=1\displaystyle y\, \left(0\right) = 1.

  1. Choose the integrating factor exp⁡ (2 t)\displaystyle \operatorname{exp}\, \left(2\, t\right), whose derivative supplies the coefficient 2.

    (e2ty)′=4e2t(e^{2t}y)'=4e^{2t}

  2. Integrate and divide by the positive factor.

    e2ty=2e2t+C,y=2+Ce−2te^{2t}y=2e^{2t}+C,\qquad y=2+Ce^{-2t}

  3. The initial value gives C=−1\displaystyle C = -1; substituting the derivative verifies the equation.

    y=2−e−2t,y′+2y=2e−2t+4−2e−2t=4y=2-e^{-2t},\qquad y'+2y=2e^{-2t}+4-2e^{-2t}=4

Conclusion: The solution tends to the equilibrium 2 as time increases.

Pitfalls & counterexamples

Tempting claim: A smooth differential equation guarantees a solution for all time.

Correction: Smoothness gives local control; the solution can still become unbounded.

Test the claim: For y prime=y squared and y (0)=1\displaystyle y\, \left(0\right) = 1, the solution is 11−t\displaystyle \frac{1}{1 - t}, whose maximal interval containing zero is t<1\displaystyle t < 1.

Check your understanding

Does dividing y prime=y (1−y)\displaystyle y\, \left(1 - y\right) by y (1−y)\displaystyle y\, \left(1 - y\right) preserve every solution?

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Compare your reasoning

No.

The constant solutions y=0\displaystyle y = 0 and y=1\displaystyle y = 1 must be recorded separately before division.

Connect this idea

The same product-rule idea underlies conservation laws and Laplace-transform solutions. Qualitative slope information can check whether a computed trajectory rises, falls, or approaches an equilibrium.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

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