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4.3 Undetermined Coefficients

idea is the same as for 2nd-order

particular solution resembles the nonhomogeneous term

throw "t" at terms that copy something.

example

\[ y''' + y'' + y' + y = e^{-t} + 4t \]

Solution: \( y = y_c + Y \)

  • \( Y \): particular (due to right side)
  • \( y_c \): complementary (right side = 0)
\[ r^3 + r^2 + r + 1 = 0 \]\[ r^2(r+1) + (r+1) = 0 \]\[ (r+1)(r^2+1) = 0 \]

\( r = -1, \quad r = -i, \quad r = i \)

Complex ALWAYS in pairs
\[ y = c_1 e^{-t} + c_2 \cos(t) + c_3 \sin(t) + Y \]
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now \( Y \): right side is \( e^{-t} + 4t \)

  • \( e^{-t} \): exponential, guess \( Ae^{-t} \)
  • \( 4t \): 1st-degree polynomial, guess \( Bt + C \)

initial \( Y \): \( Y = Ae^{-t} + Bt + C \)

copying \( c_1 e^{-t} \) fix: multiplying by \( t \)

fixed \( Y \): \( Y = Ate^{-t} + Bt + C \) (sub into the diff. eq.)

\[ Y' = -Ate^{-t} + Ae^{-t} + B \]\[ Y'' = Ate^{-t} - 2Ae^{-t} \]\[ Y''' = -Ate^{-t} + 3Ae^{-t} \]
\[ -Ate^{-t} + 3Ae^{-t} + Ate^{-t} - 2Ae^{-t} - Ate^{-t} + Ae^{-t} + B + Ate^{-t} + Bt + C = e^{-t} + 4t \]\[ 2Ae^{-t} + Bt + (B+C) = e^{-t} + 4t \]

equate coefficients

\[ 2A = 1 \implies A = \frac{1}{2} \]\[ B = 4 \]\[ B + C = 0 \implies C = -4 \]
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General Solution:

\[ y = c_1 e^{-t} + c_2 \cos(t) + c_3 \sin(t) + \frac{1}{2} t e^{-t} + 4t - 4 \]

Example: Form of \( Y \) only

\[ y^{(4)} - y''' + y'' - y' = t^2 + 4 + t \sin(t) \]
\[ r^4 - r^3 + r^2 - r = 0 \] \[ r(r^3 - r^2 + r - 1) = 0 \]

ratio of coeff: \( \frac{1}{-1} \quad \frac{1}{-1} \) — factor by grouping like last example

\[ r \left[ r^2(r-1) + (r-1) \right] = 0 \] \[ r(r-1)(r^2 + 1) = 0 \]

Roots: \( r=0, \, r=1, \, r=i, \, r=-i \)

\[ y = c_1 + c_2 e^t + c_3 \cos(t) + c_4 \sin(t) + Y \]
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right side: \( t^2 + 4 + t \sin t \)

Initial \( Y \):

\[ Y = At^2 + Bt + C + (Dt + E) \cos(t) + (Ft + G) \sin(t) \]

all get extra t because otherwise more copying

Copying \( c_1 \)

extra t for both

extra t

Fixed \( Y \):

\[ Y = t \left( At^2 + Bt + C + (Dt + E) \cos(t) + (Ft + G) \sin(t) \right) \]

Quick mention of 4.4 Variation of Parameters

4th-order

\[ y = u_1 y_1 + u_2 y_2 + u_3 y_3 + u_4 y_4 \]

still assume:

\[ u_1' y_1 + u_2' y_2 + u_3' y_3 + u_4' y_4 = 0 \]

then:

\[ u_1' y_1' + u_2' y_2' + u_3' y_3' + u_4' y_4' = 0 \] \[ u_1' y_1'' + u_2' y_2'' + u_3' y_3'' + u_4' y_4'' = 0 \] \[ u_1' y_1''' + u_2' y_2''' + u_3' y_3''' + u_4' y_4''' = g(t) \]

solve for \( u_1', u_2', u_3', u_4' \), then integrate. (Note: \( g(t) \) is the right side)

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6.1 Definition of Laplace Transform

\( f(t), \quad t \ge 0 \)

The Laplace transform of \( f(t) \) is:

Laplace transform\[ \mathcal{L} \{ f(t) \} = F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]

\( \leftarrow \) new variable

\( \uparrow \) usually upper case of the same letter

This is one example of integral transforms (another common one is Fourier transform).

What is the Laplace transform of \( f(t) = 1 \)?

\[ \mathcal{L} \{ 1 \} = F(s) = \int_{0}^{\infty} 1 \cdot e^{-st} dt \]

\( t \) is variable of integration

\( s \) is "constant" for integration

\[ = \lim_{a \to \infty} \int_{0}^{a} e^{-st} dt = \lim_{a \to \infty} \left( -\frac{1}{s} e^{-st} \Big|_0^a \right) \]
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\[ = \lim_{a \to \infty} \left( -\frac{1}{s} e^{-sa} \right) + \frac{1}{s} = \frac{1}{s} \]
\[ \mathcal{L} \{ 1 \} = \frac{1}{s}, \quad s > 0 \]

want this to go to 0

so integral converges

so, \( s > 0 \)

What about \( f(t) = \begin{cases} 1 & 0 < t < \pi \\ 0 & t \ge \pi \end{cases} \)?

Coordinate graph showing f(t) equals 1 from 0 to pi, then drops to 0 for t greater than pi.
= pi.">

discontinuous

BUT, piecewise continuous

(one of the requirements for \( f(t) \) to have a Laplace transform)

\[ \mathcal{L} \{ f(t) \} = F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]\[ = \int_{0}^{\pi} 1 \cdot e^{-st} dt + \int_{\pi}^{\infty} 0 \cdot e^{-st} dt \]\[ = -\frac{1}{s} e^{-st} \Big|_0^\pi = -\frac{1}{s} e^{-\pi s} + \frac{1}{s} = \frac{1}{s} (1 - e^{-\pi s}) \quad (s > 0) \]