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5.5 Undetermined Coeffs (continued)

Example

\[y'' - y = \cosh(x)\]

\(y = y_c + y_p\)

\(y_c\) : homogeneous only
\(y_p\) : due to right side

\(\cosh(x) = \frac{1}{2}e^x + \frac{1}{2}e^{-x}\)

\(\sinh(x) = \frac{1}{2}e^x - \frac{1}{2}e^{-x}\)

\(\frac{d}{dx} \cosh(x) = \sinh(x)\)

\(\frac{d}{dx} \sinh(x) = \cosh(x)\)

\[y_c = c_1 e^x + c_2 e^{-x} \quad \text{(solution to } y'' - y = 0\text{)}\]

for \(y_p\), there are two options

option 1: rewrite \(\cosh(x)\) as exponentials

\[y'' - y = \frac{1}{2}e^x + \frac{1}{2}e^{-x}\]

\(y_c = c_1 e^x + c_2 e^{-x}\)

\(y_p\): guess a form based on right side

\[\frac{1}{2}e^x + \frac{1}{2}e^{-x}\]

guess: \(Ae^x + Be^{-x}\)

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Check if \(y_p\) is duplicating \(y_c\)

here, both terms of \(y_p\) are duplicating \(y_c\)

fix: throw an \(x\) at \(y_p\)

\[y_p = Axe^x + Bxe^{-x}\]

now sub \(y_p\) into \(y'' - y = \frac{1}{2}e^x + \frac{1}{2}e^{-x}\)

\(y_p = Axe^x + Bxe^{-x}\)

\(y_p' = Ae^x + Axe^x + Be^{-x} - Bxe^{-x}\)

\(y_p'' = 2Ae^x + Axe^x - 2Be^{-x} + Bxe^x\)

after subbing in and equating like terms

\(A = \frac{1}{4}\)

\(B = -\frac{1}{4}\)

so, \(y_p = \frac{1}{4}xe^x - \frac{1}{4}xe^{-x}\)

and \(y = c_1 e^x + c_2 e^{-x} + \frac{1}{4}xe^x - \frac{1}{4}xe^{-x}\)

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Option 2: Work w/ Hyperbolic Functions

\[ y'' - y = \cosh(x) \]

\[ y_c = c_1 e^x + c_2 e^{-x} \]

Rewrite using hyperbolic functions

\[ \cosh(x) = \frac{1}{2} e^x + \frac{1}{2} e^{-x} \]\[ \sinh(x) = \frac{1}{2} e^x - \frac{1}{2} e^{-x} \]\[ e^x = \cosh(x) + \sinh(x) \]\[ e^{-x} = \cosh(x) - \sinh(x) \]

\[ y_c = c_1 (\cosh(x) + \sinh(x)) + c_2 (\cosh(x) - \sinh(x)) \]\[ y_c = (c_1 + c_2) \cosh(x) + (c_1 - c_2) \sinh(x) \]

Let \( c_1 + c_2 \) be represented as a new constant \( C_1 \) and \( c_1 - c_2 \) as \( C_2 \).

\[ y_c = C_1 \cosh(x) + C_2 \sinh(x) \]

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Right side: \( \cosh(x) \)

Guess \( y_p \):

\[ y_p = A \cosh(x) + B \sinh(x) \]

ALWAYS include BOTH \( \cosh(x) \) and \( \sinh(x) \) just like w/ \( \cos(x) \) and \( \sin(x) \)

Check duplication: there is \( \cosh(x) \) and \( \sinh(x) \) in \( y_c \) and \( y_p \)

Fix: multiply by \( x \)

\[ y_p = Ax \cosh(x) + Bx \sinh(x) \]\[ y_p' = A \sinh(x) + Ax \sinh(x) + B \sinh(x) + Bx \cosh(x) \]\[ y_p'' = 2A \sinh(x) + Ax \cosh(x) + Bx \sinh(x) + 2B \cosh(x) \]

Sub into \( y'' - y = \cosh(x) \), equating like terms

\[ \vdots \]\[ A = 0, \quad B = \frac{1}{2} \]\[ y_p = \frac{1}{2} x \sinh(x) \]

\[ y = C_1 \cosh(x) + C_2 \sinh(x) + \frac{1}{2} x \sinh(x) \]

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Example: Solving for Particular Solutions

\[ y^{(5)} + 2y^{(4)} - y = 3 \]

Find \( y_p \)

Guess \( y_p \): right side is 3
so, guess \( y_p = A \)

Check duplication: is there a constant in \( y_c \)?

\[ y^{(5)} + 2y^{(4)} - y = 0 \]

char. eq: \( r^5 + 2r^4 - 1 = 0 \)

how to factor??

but, if there were to be a constant in \( y_c \),
one root has to be \( r = 0 \rightarrow e^{0x} = 1 \rightarrow C \cdot 1 \)

does \( r = 0 \) work in \( r^5 + 2r^4 - 1 = 0 \)?

No, so \( r = 0 \) is NOT a root

so, there is no duplication.

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so, \( y_p = A \) is ok.

now sub it into \( y^{(5)} + 2y^{(4)} - y = 3 \)

\[ -A = 3 \] \[ \text{so, } A = -3 \]

so, \( y_p = -3 \)

Example: Finding the Form of \( y_p \)

Find the form of \( y_p \): \( y'' + y = x \cos x \)

\[ y_c = C_1 \cos x + C_2 \sin x \]

\( y_p \): guess using form of \( x \cos x \)

\[ x \cos x = (x + 0) \cos x \]

guess \( Ax + B \)

(1st-deg)

guess \( \cos(x) \) AND \( \sin(x) \)

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So, \( y_p = (Ax + B) \cos x + (Cx + D) \sin x \)

\[ = Ax \cos x + B \cos x + Cx \sin x + D \sin x \]

check for duplication: \( \cos(x) \), \( \sin(x) \) in both \( y_c \) and \( y_p \)

fix: multiply \( \cos(x) \) and \( \sin(x) \) in \( y_p \) by \( x \)

but then they duplicate \( x \cos x \) and \( x \sin x \) in \( y_p \)

fix: multiply \( \cos(x) \) and \( \sin(x) \) in \( y_p \) by \( x \) again (so, mult. by \( x^2 \))

Correct \( y_p \):

\[ y_p = Ax \cos x + Bx^2 \cos x + Cx \sin x + Dx^2 \sin x \]

or, multiply the initial guess of \( y_p \) by \( x \)

\[ y_p = Ax^2 \cos x + Bx \cos x + Cx^2 \sin x + Dx \sin x \]

Same form

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Example

\[ y^{(5)} + 2y^{(3)} + 2y'' = 3x^2 - 1 \]

form of \( y_p \)?

initial guess of \( y_p \) : \( y_p = Ax^2 + Bx + C \) (2nd deg)

Check duplication:

\[ r^5 + 2r^3 + 2r^2 = 0 \]\[ r^2 (r^3 + 2r + 2) = 0 \]

\( r = 0, 0 \) (from \( r^2 \))

not 0 (from \( r^3 + 2r + 2 \))

so, \( y_c \) has: \( y_c = C_1 + C_2x + \dots \)

\( y_p = Ax^2 + Bx + C \)

two duplication

fix: multiply entire \( y_p \) by as many factors of \( x \) until no more duplication

\( y_p = x(Ax^2 + Bx + C) \) still duplicating parts of \( y_c \)

fix: another \( x \)

\( y_p = x^2(Ax^2 + Bx + C) \) now no more duplication!