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Finish exercices equa diff ordre 1

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flyingscorpio@arch-desktop 2021-09-19 19:57:14 +02:00
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52
analyse/exercices/main.tex
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@ -33,8 +33,12 @@
y_1' = 2ax + b \\
\end{array}
\right.
\quad y_1' - 2 y_1 = x^2 \Leftrightarrow
\end{align*}
\begin{align*}
y_1' - 2 y_1 = x^2
&\Leftrightarrow
2ax + b - 2(ax^2 + bx + c) = x^2 \\
&\Leftrightarrow
\left\{
\begin{array}{l}
-2a = 1 \\
@ -49,12 +53,13 @@
b = \sfrac{-1}{2} \\
c = \sfrac{-1}{4} \\
\end{array}
\right. \\
\right.
\end{align*}
\begin{align*}
\implies y_1 = -\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4}
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{-\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4} + \lambda e^{2x}}
\end{equation*}
@ -80,17 +85,20 @@
y_1' = ae^{3x} + 3axe^{3x} = (3ax + a)e^{3x} \\
\end{array}
\right.
\quad 3y_1' - 9y_1 = 7e^{3x}
\end{align*}
\begin{align*}
3y_1' - 9y_1 = 7e^{3x}
&\Leftrightarrow 3(3ax + a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
&\Leftrightarrow (9ax + 3a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
&\Leftrightarrow 9ax + 3a - 9ax = 7 \\
&\Leftrightarrow 3a = 7 \\
&\Leftrightarrow a = \frac{7}{3} \\
&\Leftrightarrow a = \frac{7}{3}
\end{align*}
\begin{align*}
\implies y_1 = \frac{7}{3}xe^{3x}
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \lambda e^{3x} + \frac{7}{3}xe^{3x} = \boxed{(\frac{7}{3}x + \lambda)e^{3x}}
\end{equation*}
@ -116,15 +124,18 @@
y_1' = 3ae^{3x} \\
\end{array}
\right.
\quad 2y_1' - 4y_1 = 5e^{3x}
\end{align*}
\begin{align*}
2y_1' - 4y_1 = 5e^{3x}
&\Leftrightarrow 6ae^{3x} - 4ae^{3x} = 5e^{3x} \\
&\Leftrightarrow 2a = 5 \\
&\Leftrightarrow a = \frac{5}{2} \\
&\Leftrightarrow a = \frac{5}{2}
\end{align*}
\begin{align*}
\implies y_1 = \frac{5}{2}e^{3x}
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{\lambda e^{2x} + \frac{5}{2}e^{3x}}
\end{equation*}
@ -168,8 +179,29 @@
4a - 8b = 0 \\
\end{array}
\right.
% TODO: finish
\Leftrightarrow
\left\{
\begin{array}{l}
10b = 3 \\
4b - 2a = 0 \\
\end{array}
\right. \\
&\Leftrightarrow
\left\{
\begin{array}{l}
b = \sfrac{3}{10} \\
a = \sfrac{12}{20} = \sfrac{3}{5} \\
\end{array}
\right.
\end{align*}
\begin{align*}
\implies y_1 = \frac{3}{5}\cos(2x) + \frac{3}{10}\sin(2x)
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{\lambda e^{-4x} + \frac{3}{5}\cos(2x) + \frac{3}{10}\sin(2x)}
\end{equation*}
\end{enumerate}