Solving differential equations



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Example

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).

Solving linear equations

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