John Travis
Mississippi College
Solve $y'' + 9y = \sin(3t)$
$(D^2 + 1) y = \sin(3t)$
yields solutions for the homogeneous part of
$y_1 = \sin(3t)$ and $y_2 = \cos(3t)$.
Operating on both sides with operator $D^2 + 1$ makes the original DE also homogeneous.
$(D^2 + 1)^2 y = (D^2+1)\sin(3t) = 0$
Thus, we have new solutions $y_3 = t \sin(3t)$ and $y_4 = t \cos(3t)$.
Therefore, try the particular solution
$y_p = A t \sin(3t) + B t \cos(3t)$.
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\newcommand{\Bold}[1]{\mathbf{#1}}6 \, A \cos\left(3 \, t\right) - 6 \, B \sin\left(3 \, t\right) = \sin\left(3 \, t\right)
\newcommand{\Bold}[1]{\mathbf{#1}}6 \, A \cos\left(3 \, t\right) - 6 \, B \sin\left(3 \, t\right) = \sin\left(3 \, t\right)
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Solving this using our own heads yields:
$A = 0$ and $B = -1/6$.
Let's check the general solution:
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\newcommand{\Bold}[1]{\mathbf{#1}}\sin\left(3 \, t\right) = \sin\left(3 \, t\right) True \newcommand{\Bold}[1]{\mathbf{#1}}\sin\left(3 \, t\right) = \sin\left(3 \, t\right) True |
Below, you can check your own differential equation by changing the appropriate parts from above and letting Sage take care of the nitty gritty details....
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\newcommand{\Bold}[1]{\mathbf{#1}}4 \, A e^{\left(3 \, t\right)} = 5 \, e^{\left(3 \, t\right)}
\newcommand{\Bold}[1]{\mathbf{#1}}4 \, A e^{\left(3 \, t\right)} = 5 \, e^{\left(3 \, t\right)}
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\newcommand{\Bold}[1]{\mathbf{#1}}5 \, e^{\left(3 \, t\right)} = 5 \, e^{\left(3 \, t\right)} YES! \newcommand{\Bold}[1]{\mathbf{#1}}5 \, e^{\left(3 \, t\right)} = 5 \, e^{\left(3 \, t\right)} YES! |
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