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Second-order, constant coefficients

ay″ + by′ + cy = 0 and the characteristic equation: oscillation, damping, growth.

Try y = e^{rx} and the equation becomes a quadratic in r, the characteristic equation. Real distinct roots give exponentials, a repeated root adds an x, complex roots give sines and cosines. y″ + y = 0 is the simple harmonic oscillator: every pendulum, spring and LC circuit.

Przykład pracownika · y'' + y = 0

Differential-Equations · Differential-Equations

Y'' + y = 0

y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 0

Krok po kroku

  1. y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 0

    The differential equation.

  2. \text{order } 2

    Order 2: the highest derivative present.

  3. \text{Constant coefficients, homogeneous}

    Try y = e^{rx}: the characteristic polynomial in r gives the roots, and each root contributes a term to the general solution.

  4. r^{2} + 1 = 0

    Characteristic equation of the homogeneous part (substitute y = e^{rx}).

  5. r = - i,\ r = i

    Its roots.

  6. y{\left(x \right)} = C_{1} \sin{\left(x \right)} + C_{2} \cos{\left(x \right)}

    General solution (C₁, C₂ … are arbitrary constants).

  7. \checkmark

    Verified: substituting the solution back into the equation gives 0.

Odkryj odpowiedź
y{\left(x \right)} = C_{1} \sin{\left(x \right)} + C_{2} \cos{\left(x \right)}

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Jak to zrobić?: Second-order, constant coefficients

  1. The differential equation.
  2. Order 2: the highest derivative present.
  3. Try y = e^{rx}: the characteristic polynomial in r gives the roots, and each root contributes a term to the general solution.
  4. Characteristic equation of the homogeneous part (substitute y = e^{rx}).
  5. Its roots.
  6. General solution (C₁, C₂ … are arbitrary constants).
  7. Verified: substituting the solution back into the equation gives 0.

Pytania, które ludzie zadają

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

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