10.2.1 Numerical Methods in Context

Non-Homogeneous Second-Order Differential Equation (S.O.D.E)

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1. Solve the characteristic equation: The characteristic equation is derived by assuming a solution of the form $y = e^{mx}$, leading to:

$$m^2 - 5m + 6 = 0$$

Factorizing the quadratic:

$$(m - 2)(m - 3) = 0$$

This gives the roots $m = 2$ and $m = 3$.

2. Write the general solution to the homogeneous equation: The general solution (complementary function) is formed by the linear combination of the solutions corresponding to the roots:

$$y_h = Ae^{2x} + Be^{3x}$$

where $A$ and $B$ are arbitrary constants.

3. Combine with the particular integral: From the previous work, the particular integral (the particular solution to the non-homogeneous equation) was found to be:

$$y_p = -\frac{1}{6} \sin(3x) + \frac{5}{6} \cos(3x)$$

4. Form the complete solution: The complete solution to the non-homogeneous differential equation is the sum of the particular integral and the complementary function:

$$y = -\frac{1}{6} \sin(3x) + \frac{5}{6} \cos(3x) + Ac^{2x} + Bc^{3x}$$

5. Note on the complementary function: The complementary function on its own is not a solution to the original non-homogeneous equation (since it equals zero), but it complements the particular integral to form the full solution.

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Problem Statement

1. Complementary Function (C.F.)

First, solve the homogeneous part of the equation:

$$\text{C.F.:} \quad m^2 - 5m + 6 = 0$$

Factorizing:

$$(m - 2)(m - 3) = 0$$

This gives the roots $m = 2$ and $m = 3$.

Thus, the complementary function is:

$$y_h = Ae^{2x} + Be^{3x}$$

2. Particular Integral (P.I.)

Next, propose a form for the particular integral. Since the non-homogeneous term is $3x^2$, assume:

$$y = ax^2 + bx + c$$

Then:

$$\frac{dy}{dx} = 2ax + b$$ $$\frac{d^2y}{dx^2} = 2a$$

Substitute $y$ and its derivatives into the differential equation:

$$2a - 5(2ax + b) + 6(ax^2 + bx + c) = 3x^2$$

Simplifying, collect terms involving $x^2, x,$ and constants:

$$6ax^2 + (6b - 10a)x + (2a - 5b + 6c) = 3x^2$$

Equating coefficients of like terms:

• Coefficient of $x^2: 6a = 3 ⟹ a = \frac{1}{2}$
• Coefficient of $x: 6b - 10\left(\frac{1}{2}\right) = 0 ⟹ b = \frac{5}{6}$
• Constant term: $2\left(\frac{1}{2}\right) - 5\left(\frac{5}{6}\right) + 6c = 0 ⟹ c = \frac{19}{36}$ So, the particular integral is:

$$y_p = \frac{1}{2}x^2 + \frac{5}{6}x + \frac{19}{36}$$

3. General Solution

The general solution is the sum of the complementary function and the particular integral:

$$\boxed {y = \frac{1}{2}x^2 + \frac{5}{6}x + \frac{19}{36} + Ae^{2x} + Be^{3x}}$$

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Forms of Particular Integrals

RHS	P.I.
k	c
$\lambda x+ \mu$	$bx+c<br>$ Note: $\mu$ can = 0
$\lambda x^2+ \mu x+2$	$ax^2+bx+c$ Note: $\mu$, 2 can = 0
$\lambda e^{kx}$	$ae^{kx}$
$p\sin \lambda x+q\cos \lambda x$	$a\sin \lambda x+b\cos \lambda x$ Note: $a$,$b$ can = 0

Anything more complemented will be given in the question.

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S.O.D.E. Where P.I. is Part of C.F.

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Justification of Trying $\lambda x e^{2x}$:

1. Start with the given S.O.D.E.:

$$\frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = e^{2x}$$

2. Rewrite the differential equation:

• This can be factored as:

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

• Multiply both sides by $\left(\frac{d}{dx} - 2\right)$ to get:

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

3. Calculate the right-hand side:

• Simplify the expression on the right:

$$\left(\frac{d}{dx} - 2\right)e^{2x} = \frac{d}{dx} e^{2x} - 2e^{2x} = 2e^{2x} - 2e^{2x} = 0$$

• This results in:

$$\left(\frac{d}{dx} - 2\right)^2\left(\frac{d}{dx} - 3\right)y = 0$$

4. Conclusion:

• This leads to the equation:

$$\left(\frac{d}{dx} - 2\right)^2\left(\frac{d}{dx} - 3\right)y = 0$$

• The repeated root at $m = 2$ results in the solution of the form:

$$y = (A + Bx)e^{2x}$$

• Here, the term $Bx e^{2x}$ is equivalent to the form $\lambda x e^{2x}$ that was tried, justifying its use as a trial solution.

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