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Nonhomogeneous equations

Driving forces: homogeneous solution plus a particular solution.

With a forcing term on the right, the general solution is the homogeneous solution plus any one particular solution. Undetermined coefficients guesses the particular solution in the shape of the forcing (exponential, polynomial, sinusoid); variation of parameters handles the rest.

Kushanda muenzaniso · y'' - 3y' + 2y = e^x

Y'' - 3y' + 2y = e^x

2 y{\left(x \right)} - 3 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = e^{x}

Step by step

  1. 2 y{\left(x \right)} - 3 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = e^{x}

    The differential equation.

  2. \text{order } 2

    Order 2: the highest derivative present.

  3. \text{Undetermined coefficients}

    Solve the homogeneous part from the characteristic equation, then guess a particular solution shaped like the right-hand side.

  4. r^{2} - 3 r + 2 = 0

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

  5. r = 2,\ r = 1

    Its roots.

  6. y{\left(x \right)} = \left(C_{1} + C_{2} e^{x} - x\right) e^{x}

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

  7. \checkmark

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

Kuratidza mhinduro
y{\left(x \right)} = \left(C_{1} + C_{2} e^{x} - x\right) e^{x}

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Maitiro: Nonhomogeneous equations

  1. The differential equation.
  2. Order 2: the highest derivative present.
  3. Solve the homogeneous part from the characteristic equation, then guess a particular solution shaped like the right-hand side.
  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.

Mabvunzo vanhu vanobvunza

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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