hsf/ma5/notes/test/test.tex

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		elalmqvist@gmail.com --- \url{https://wych.dev}
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\begin{document}

\title{Anteckningar 2022-04-28}
\author{Elias Almqvist}

\maketitle
\newpage

Lös ekvationen:
$$
y'' = -2y' - 8y + \sin x
$$

Lösning:
$$
y'' = -2y' - 8y + \sin x, \quad + 2y' + 8y
$$
$$
y'' + 2y' + 8y = \sin x
$$

Vi får därmed att $y = y_p + y_h$, \\
Börjar med den homogena:
$$
y'' + 2y' + 8y = 0
$$
$$
\implies r^2 + 2r + 8 = 0 \quad \text{antag att y är } e^{rx}
$$
$$
\implies r = -1 \pm \sqrt{-7} = -1 \pm 7i
$$
$$
\therefore y_h = e^{-x} \left( A\cos(\sqrt{7}x) + B\sin{\sqrt{7}x} \right) \quad | \quad A,B \in \mathbb{R}
$$

Part lösn: antar att $y_p$ är av formen: $y_p = D\sin x + E\cos x + F \quad | \quad D,E,F \in \mathbb{R}$ \\
Stoppar in den i ekvationen:
$$
y_p' = D\cos x - E\sin x 
$$
$$
y_p'' = -\left( D\sin x + E\cos x \right)
$$
Vi får därmed:
$$
-\left( D\sin x + E\cos x \right) + 2\left(D\cos x - E\sin x\right) + 8\left(D\sin x + E\cos x + F\right) = \sin x
$$
$$
-D\sin x - E\cos x + 2D\cos x - 2E\sin x + 8D\sin x + 8E\cos x + 8F = \sin x
$$
$$
2D\cos x - 2E\sin x + 7D\sin x + 7E\cos x + 8F = \sin x
$$
$$
\implies
\begin{cases}
	8F = 0 \implies F = 0 \\
	7E + 2D = 0 \\
	7D - 2E = 1
\end{cases}
$$

Stoppar in i en matris och får:
$$
\begin{bmatrix}
	7 & 2 & 0 \\
	-2 & 7 & 1
\end{bmatrix}
\rightarrow
\begin{bmatrix}
	1 & 0 & -\frac{2}{53} \\
	0 & 1 & \frac{7}{53}
\end{bmatrix}
$$
$$
\therefore y_p = \frac{7}{53} \sin x - \frac{2}{53} \cos x
\quad \because  E = -\frac{2}{53}, \quad D = \frac{7}{53}
$$

Slutligen får vi därmed hela allmäna lösningen:
$$
\therefore y = y_p + y_h 
$$
$$
= \frac{7}{53} \sin x - \frac{2}{53} \cos x + e^{-x} \left( A\cos(\sqrt{7}x) + B\sin{\sqrt{7}x} \right)
$$

Givet att $A,B \in \mathbb{R}$

\end{document}