diff --git a/analyse/exercices/main.tex b/analyse/exercices/main.tex index 17fb9bd..ddbf208 100644 --- a/analyse/exercices/main.tex +++ b/analyse/exercices/main.tex @@ -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}