Solving differential equations



		Site Map

Linear ODEs with variable coefficient

Natural oscillations (be it mechnical or electrical circuit) exhibit a forcing function that is due to friction, dashpot, or circuit resistance.

Suppose we model this forcing function as $f (t)$ , an linear ODE with this added nonhomogeneous term now takes the form

$A_n \frac{{d^n y}}{{dt^n }} + A_{n - 1} \frac{{d^{n - 1} y}}{{dt^{n - 1} }} + \cdots + A_1 \frac{{dy}}{{dt}} + A_0 y = f\left( t \right),$

or simply (in standard form),

$a_n y^{(n)} + a_{n - 1} y^{(n - 1)} + \cdots + a_1 y' + a_0 y = f\left( t \right).\,$

In case of non-homogeneous linear ODE (non-HLDE) where the input function is polynomial, sinusodial, exponential or any product of the three; we seek the solution to the equation above in the form of $y G = y c + y p$ where

$y G$ denotes a general solution;
$y c$ denotes a characteristic equation;
$y p$ denotes a particular solution.

Method of undetermined coefficients

The method of undetermined coefficients (MoUC) is useful in finding solution for $y p$ . Given $P (D) = f (t)$ , find the annihilator $A (D)$ for $f (t)$ such that $A (D) f (t) = 0$ ; then apply $A (D)$ to both side of $P (D) = f (t)$ to have $A (D) f (t) = A (D) f = 0$ , a HLDE with constant coefficients (cc) which could than be readily solve using technique found in 3.1. Note by convention when f(t) is used, it often means that an equation is time-dependent, where f(x) and other denotes time-independent.

Suppose that f(x) = 1 − 2x; A(D) has the following family of solutions:

Recall: $r = 0: e 0 = 1, x, x 2, x 3,...$

Thus, when we have x; henceforth it implies this root repeated twice. With this in mind, $A (D) = D 2$ has multiplicity 2.

Similarly, case of complex roots is based on sin or cos.

Example: $f (x) = sin x - x cos2 x$

sin x is due to complex root, has real part of 0 because $e 0 = 1$ (multiply 1 on sin and cos).
A(D) then has root of $0 \pm i$ (simply $\pm i$ ) with multiplicity 1.
Also $r=\pm 2i$ with multiplicity 2.

Example: $\left[ {D^2 - D} \right]y = 1 - 2x$

Here $r = 0: e 0 = 1, x, x 2,.... x n; r = 1: e x, x e x,..., x n e x$ Note that once a distinct root is used, it may not be used again due to linearly independent.

$y_c = c_1 y_1 + c_2 y_2 = c_1 \left( 1 \right) + c_2 \left( {e^x } \right)$ . A(D) has of multiplicity of 2.

$\left. {\begin{matrix} {Y_p = Ax + Bx^2 } \\ {Y_p ^\prime = A + 2Bx} \\ {Y_p ^{\prime \prime } = 2B} \\ \end{matrix}} \right\}2B - \left[ {A + 2Bx} \right] = \left[ {2B - A} \right] - 2Bx = 1 - 2x$

Equating coefficients, 2B − A yields constant term on RHS of 1, hence 2B − 1 = 1 so B = 1, A = 1. −2B = −2. Therefore $y p = A x + B x 2 = x + x 2$ . Solution hence becomes $y = y c + y p = C 1 + C 2 e x + x + x 2$ . If we do not keep deleting our used roots, we than may have $y = y c + y p = C 1 + C 2 e x + 1 + x 2$ , it would be incorrect since C₁ absorbs the arbitrariness of x (here is 1); thus violates linearly dependence.

Example: $\left[ {D^2 - D} \right]y = x - 2e^x$ (same as $y'' - y' = x - 2 e x$ )

In this case, we have roots r = {0, 1} which yield family of solution such as

$\begin{matrix} r = 0:1,x,x^2 ,x^3 ,... \\ r = 1:e^x ,xe^x ,x^2 e^x ,... \\ \end{matrix}$

Therefore, $y 1 = 1, y 2 = e x$ and $y c = C 1 (1) + C 2 e x$ Since A(D) has $\left. \begin{matrix} r = 0\,\,{\rm{of\ multiplicity\ of\ 2}} \\ r = 1\,\,{\rm{of\ multiplicity\ of\ 1}} \\ \end{matrix} \right\}$ giving the form of $\left. \begin{matrix} Y_p = Ax + Bx + Cxe^x \\ Y_p ^\prime = A + 2Bx + C(1 + x)e^x \\ Y_p ^{\prime \prime } = 2B + C(2 + x)e^x \\ \end{matrix} \right\}$ put in original equation to have $\left[ {2B - A} \right] - 2Bx + ce^x = x - 2e^x$

Equating coefficient, $\begin{matrix} 2B - A = 0\,\,{\rm{so }}\,{\rm{A = 2B}} \Rightarrow A = - 1 \\ - 2B = 1 \Rightarrow B = - \frac{1}{2};C = - 1 \\ \end{matrix}$

Thus $y_p = Ax + Bx + Cxe^x = - x - \frac{1}{2}x^2 - 2xe^x$

Example: $\left[ {D^2 + 1} \right]y = f = \sec x$ . What roots would give rise to the solution of the form $f\left( x \right) = \sec \left( x \right)$ ?

Solution: No roots. $f\left( x \right) = \sec \left( x \right)$ is not a sinusoid, rather the reciprocal of a sinusoid. So this method would not apply and 2nd-order variation-of-parameters (VoP) must be used to solve these type of problems (no valid finite linear combination could be tried in this case).

Method of variation of parameters

As explained above, the general solution to a non-homogeneous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ can be expressed as the sum of the general solution $y h (x)$ to the corresponding homogenous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = 0$ and any one solution $y p (x)$ to $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ .

Like the method of undetermined coefficients, described above, the method of variation of parameters is a method for finding one solution to $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ , having already found the general solution to $y''(x) + p (x) y'(x) + q (x) y (x) = 0$ . Unlike the method of undetermined coefficients, which fails except with certain specific forms of g(x), the method of variation of parameters will always work; however, it is significantly more difficult to use.

For a second-order equation, the method of variation of parameters makes use of the following fact:

Fact

Let p(x), q(x), and g(x) be functions, and let $y 1 (x)$ and $y 2 (x)$ be solutions to the homogeneous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = 0$ . Further, let u(x) and v(x) be functions such that $u'(x) y 1 (x) + v'(x) y 2 (x) = 0$ and $u'(x) y 1'(x) + v'(x) y 2'(x) = g (x)$ for all x, and define $y p (x) = u (x) y 1 (x) + v (x) y 2 (x)$ . Then $y p (x)$ is a solution to the non-homogeneous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ .

Proof

$y p (x) = u (x) y 1 (x) + v (x) y 2 (x)$

$y p'(x) = u'(x) y 1 (x) + u (x) y 1'(x) + v'(x) y 2 (x) + v (x) y 2'(x) = 0 + u (x) y 1'(x) + v (x) y 2'(x)$

$y p''(x) = u'(x) y 1'(x) + u (x) y 1''(x) + v'(x) y 2'(x) + v (x) y 2''(x) = g (x) + u (x) y 1''(x) + v (x) y 2''(x)$

$y p''(x) + p (x) y' p (x) + q (x) y p (x) = g (x) + u (x) y 1''(x) + v (x) y 2''(x) + p (x) u (x) y 1'(x) + p (x) v (x) y 2'(x) + q (x) u (x) y 1 (x) + q (x) v (x) y 2 (x) = g (x) + u (x)(y 1''(x) + p (x) y 1'(x) + q (x) y 1 (x)) + v (x)(y 2''(x) + p (x) y 2'(x) + q (x) y 2 (x)) = g (x) + 0 + 0 = g (x)$

Usage

To solve the second-order, non-homogeneous, linear differential equation $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ using the method of variation of parameters, use the following steps:

Find the general solution to the corresponding homogeneous equation $y''(x) + p (x) y'(x) + q (x) y (x) = 0$ . Specifically, find two linearly independent solutions $y 1 (x)$ and $y 2 (x)$ .
Since $y 1 (x)$ and $y 2 (x)$ are linearly independent solutions, their Wronskian $y 1 (x) y 2'(x) - y 1'(x) y 2 (x)$ is nonzero, so we can compute $\frac{g(x) y_2(x)}{y_1(x) y_2'(x) - y_1'(x) y_2(x)}$ and $-\frac{g(x) y_1(x)}{y_1(x) y_2'(x) - y_1'(x) y_2(x)}$ . If the former is equal to u'(x) and the latter to v'(x), then u and v satisfy the two constraints given above: that $u'(x) y 1 (x) + v'(x) y 2 (x) = 0$ and that $u'(x) y 1'(x) + v'(x) y 2'(x) = g (x)$ .
Integrate $\frac{g(x) y_2(x)}{y_1(x) y_2'(x) - y_1'(x) y_2(x)}$ and $-\frac{g(x) y_1(x)}{y_1(x) y_2'(x) - y_1'(x) y_2(x)}$ to obtain u(x) and v(x), respectively. (Note that we only need one choice of u and v, so there is no need for constants of integration.)
Compute $y p (x) = u (x) y 1 (x) + v (x) y 2 (x)$ . The function $y p$ is one solution of $y''(x) + p (x) y'(x) + q (x) y (x) = g (x)$ .
The general solution is $c 1 y 1 (x) + c 2 y 2 (x) + y p (x)$ , where $c 1$ and $c 2$ are arbitrary constants.

Higher-order equations

The method of variation of parameters can also be used with higher-order equations. For example, if $y 1 (x)$ , $y 2 (x)$ , and $y 3 (x)$ are linearly independent solutions to $y'''(x) + p (x) y''(x) + q (x) y'(x) + r (x) y (x) = 0$ , then there exist functions u(x), v(x), and w(x) such that $u'(x) y 1 (x) + v'(x) y 2 (x) + w'(x) y 3 (x) = 0$ , $u'(x) y 1'(x) + v'(x) y 2'(x) + w'(x) y 3'(x) = 0$ , and $u'(x) y 1''(x) + v'(x) y 2''(x) + w'(x) y 3''(x) = g (x)$ . Having found such functions (by solving algebraically for u'(x), v'(x), and w'(x), then integrating each), we have $y p (x) = u (x) y 1 (x) + v (x) y 2 (x) + w (x) y 3 (x)$ , one solution to the equation $y'''(x) + p (x) y''(x) + q (x) y'(x) + r (x) y (x) = g (x)$ .

Example

Solve the previous example, $y'' + y = sec x$ Recall $\sec x = \frac{1}{{\cos x}} = f$ . From technique learned from 3.1, LHS has root of $r = \pm i$ that yield $y c = C 1 cos x + C 2 sin x$ , (so $y 1 = cos x$ , $y 2 = sin x$ ) and its derivatives $\left\{ {\begin{matrix} {\dot u = \frac{{ - y_2 f}}{W} = \frac{{ - \sin x}}{{\cos x}} = \tan x} \\ {\dot v = \frac{{y_1 f}}{W} = \frac{{\cos x}}{{\cos x}} = 1} \\ \end{matrix}} \right.$ where Wronskian $W\left( {y_1 ,y_2 :x} \right) = \left| {\begin{matrix} {\cos x} & {\sin x} \\ { - \sin x} & {\cos x} \\ \end{matrix}} \right| = 1$ were computed in order to seek solution to its derivatives. Upon integration, $\left\{ \begin{matrix} u = - \int {\tan xdx = - \ln \left| {\sec x} \right| + C} \\ v = \int {1dx = x + C} \\ \end{matrix} \right.$ Computing $y p$ and $y G$ : $\begin{matrix} y_p = f = uy_1 + vy_2 = \cos x\ln \left| {\cos x} \right| + x\sin x \\ y_G = y_c + y_p = C_1 \cos x + C_2 \sin x + x\sin x + \cos x\ln \left( {\cos x} \right) \\ \end{matrix}$

Solving linear equations

What do you imagine thinking about rules for baccarat? I know.; Do you know where can I find online craps games online? I see:); Looking at no deposit bonus bingo you could see how to get money for free.; Highest jackpots all at the most trusted gambling place, the game play roulette!; Read all the penny slots online monthly updated articles and detect the best gambling room!; generic drug lexapro; buy clomid; generic inderal; guaranteed cheapest viagra