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II Order Differential Equations with Constant Coefficients

Posted in differential, equation by beauangel on the February 17, 2007

Standard Form

$a_0 \frac{d^2y}{dx^2} + a_1 \frac{dy}{dx} + a_2y = f(x)$ where $a_0, a_1, a_2$ are constants.

The general solution of such a differential equation contains two parts; the Complementary Function and the Particular Integral.

To find the Complementary Function:

Write the auxiliary equation in the form: $a_0m^2 + a_1m + a_2 =0$ where the differential equation is in the standard form.

Solving the auxiliary equation, a quadratic, will yield two roots, say m₁ and m₂.

The Complementary Function (CF) is then written based on the roots.

If m₁ and m₂ are real and different, then the CF is $y = Ae^{m_1x} + Be^{m_2x}$
If m₁ and m₂ are real and equal, ie m₁ = m₂ = m, then the CF is $y = (A + Bx)e^{mx}$
If m₁ and m₂ are complex roots of the form $\alpha + i\beta$ and $\alpha - i\beta$ , then the CF is $y = e^{\alpha x} (A \cos \beta x + B \sin \beta x)$ where $A$ and $B$ are arbitrary constants.

The Particular Integral is similar to the function of x, $f(x)$ on the RHS of the differential equation and it is determined by the method of undetermined coefficients.

If the RHS of the equation is a constant, then we use a trial PI of K (a different constant). A trial PI of a general first degree form $A_0x +A_1$ and a general second degree form $A_0x^2 + A_1x + A_2$ is used for functions of $ax + b$ and $ax^2 + bx +c$ respectively. If the function is of the form $be^{ax}$ , then we use $Ke^{ax}$ , the same exponential function. Similarly, trigonometric functions will have trial PIs of the form $L \cos bx + K \sin bx$ .

$K, L, A_0, A_1, A_2$ are undetermined coefficients. By finding $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ and substituting $y, \frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the given differential equation, the values of the undetermined coefficients can be obtained. Hence the Particular Integral is found.

Combine the Complementary Function and the Particular Integral and we will get the General Solution of the Second Order Differential Equation with constant coefficients.

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I Order Differential Equations Part II

Posted in differential, equation by beauangel on the February 17, 2007

Linear Differential Equations

Standard Form:

$\frac{dy}{dx} + p\:y = Q(x)$ where $p$ and $Q$ are functions of $x$ .

To solve this type of differential equation, we multiply the equation throughout by the factor $e^{\int p dx}$ called the Integrating Factor and hence convert the LHS into an exact differential.

Multiplying throughout by $e^{\int p dx}$ , we get

$e^{\int p dx}. \frac{dy}{dx} + e^{\int p dx}.p\:y = Q\:e^{\int p dx}$

which becomes

$\frac{d}{dx} \left( e^{\int p dx}y \right) = Qe^{\int p dx}$

Integrating both sides w.r.t x, we get

$\int \frac{d}{dx} = \int Q e^{\int p dx}dx$

ie, $e^{\int p dx}y = \int Q e^{\int p dx}dx$

Therefore, the formula for the General Solution of a Linear Differential Equation is:

$ye^{\int p dx} = \int Q e^{\int p dx}dx$

I Order Differential Equations Part I

Posted in differential, equation by beauangel on the February 16, 2007

Standard Form

$\frac{dy}{dx} = f(x).g(x)$

To solve:

Make sure that the given differential equation is in the standard form of a variable separable equation.
Collect the function of x and the differential $dx$ to one side and the function of y and $dy$ to the other side of the differential equation.
Integrate both sides with respect to the corresponding variables and get the solution. Add the constant of integration to one side.