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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.

Намунаи коркардашуда · 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}

Қадами ба қадам

  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.

Ҷавоби ҷавобро нишон диҳед
y{\left(x \right)} = \left(C_{1} + C_{2} e^{x} - x\right) e^{x}

Ҳоло шумо Масъаларо интихоб кунед, ё ки онро навишта ё кашед. Ҳар қадам, тасвир, ҷавоб то он даме, ки шумо пурсед, пинҳон мешавад.

Кор кардан

Дар ҳисоби ройгон ба ҳар як дарс қайдҳо илова карда мешаванд, қайди он, ки шумо чӣ кор кардаед, масъалаҳои ҳалшуда дар як ҷо ва муаллиме, ки шумо метавонед дар бораи ин саҳифа пурсед. Математика барои ҳама кушода аст, хоҳ ворид шуда бошад ё на.

Бақайдгирӣ Ворид шудан

Символы, используемые здесь

Барои муайянкунии пурраи маъно, тасвир ва маънои ҳар як ҳарф дар он, ба ҳар як аломат пахш кунед.

Чӣ тавр: 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.

Саволҳои маъмул

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.

& Иловаи забон Differential Equations