Ојлерова диференцијална једначина
\[\begin{array} \\ {A_n}{\left( {ax + b} \right)^n}{y^{\left( n \right)}} + {A_{n - 1}}{\left( {ax + b} \right)^{n - 1}}{y^{\left( {n - 1} \right)}} + ... + {A_1}\left( {ax + b} \right)y' + {A_0}y = f\left( x \right) \\ {A_n},{A_{n - 1}},...,{A_1},{A_0},a,b - const. \\ ax + b > 0 \\ \\ {\text{ смена : }} ax + b = {e^t} \\ {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} {\text{ }}{\text{ }} t = \ln \left( {ax + b} \right) \\ {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} {\text{ }}{\text{ }} dt = \frac{1}{{ax + b}} \cdot adx \\ {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} {\text{ }}{\text{ }} \frac{{dt}}{{dx}} = {e^{ - t}} \cdot a \\ \\ y' = \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} = \dot y \cdot {e^{ - t}}a \\ y'' = \frac{{dy'}}{{dx}} = \frac{{dy'}}{{dt}} \cdot \frac{{dt}}{{dx}} = \frac{d}{{dt}}\left( {\dot y \cdot {e^{ - t}}a} \right) \cdot {e^{ - t}}a = \\ {\text{ }} {\text{ }} {\text{ }} {\text{ }}= \left( {\ddot y \cdot {e^{ - t}}a + \dot y \cdot \left( { - {e^{ - t}}a} \right)} \right) \cdot {e^{ - t}}a = {a^2}{e^{ - 2t}}\left( {\ddot y - \dot y} \right) \end{array}\]
Задаци
1. Реши диференцијалну једначину ${x^2}y'' + 2xy' - 2y = 2{x^2} + 3\ln x + 2$.
Препознајемо да је дата диференцијална једначина Ојлерова, па уводимо смену:
$x = {e^t}{\text{ }}{\text{ }} {\text{ }} \Rightarrow {\text{ }}{\text{ }} {\text{ }} t = \ln x$
$dx = {e^t}dt$
$\frac{{dt}}{{dx}} = {e^{ - t}}$
$y' = \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} = \dot y \cdot {e^{ - t}}$
$y'' = \frac{{dy'}}{{dx}} = \frac{{dy'}}{{dt}} \cdot \frac{{dt}}{{dx}} = \frac{d}{{dt}}\left( {\dot y \cdot {e^{ - t}}} \right) \cdot {e^{ - t}} = $
$ {\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} = \left( {\ddot y{e^{ - t}} + \dot y\left( { - {e^{ - t}}} \right)} \right) \cdot {e^{ - t}} = \left( {\ddot y - \dot y} \right) \cdot {e^{ - 2t}}$
Уврштавајући у полазну једначину добијамо
$\left( {\ddot y{e^{ - t}} + \dot y\left( { - {e^{ - t}}} \right)} \right) \cdot {e^{ - t}} = \left( {\ddot y - \dot y} \right) \cdot {e^{ - 2t}}$
${e^{2t}}\left( {\ddot y - \dot y} \right) \cdot {e^{ - 2t}} + 2{e^t}\dot y \cdot {e^{ - t}} - 2y = 2{e^{2t}} + 3t + 2$
$\ddot y - \dot y + 2\dot y - 2y = 2{e^{2t}} + 3t + 2$
$\ddot y + \dot y - 2y = 2{e^{2t}} + 3t + 2$
Овим смо нашу диференцијалну једначину свели на диференцијалну једначину са константним коефицијентима, коју даље лако решавамо.
Најпре решавамо хомогени део:
${k^2} + k - 2 = 0$
${k_1} = - 2{\text{ }}{\text{ }} \wedge {\text{ }}{\text{ }} {k_2} = 1$
${y_h} = {C_1}{e^{ - 2t}} + {C_2}{e^t}$
Враћајући смену $t = \ln x$ добијамо да је решење хомогеног дела
${y_h} = \frac{{{C_1}}}{{{x^2}}} + {C_2}x$
${f_1}\left( x \right) = 2{e^{2t}}$
$\left. \begin{gathered}
a = 2 \hfill \\
b = 0 \hfill \\
\end{gathered} \right\} \Rightarrow {\text{ }}a \pm bi = 2{\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}r = 0$
$\left. \begin{gathered}
n = 0 \hfill \\
m = 0 \hfill \\
\end{gathered} \right\} \Rightarrow {\text{ }} {\text{ }}N = 0$
${y_{{p_1}}} = {t^0}{e^{2t}}\left( {A\cos 0t + B\sin 0t} \right)$
${y_{{p_1}}} = A{e^{2t}}$
${{\dot y}_{{p_1}}} = 2A{e^{2t}}$
${{\ddot y}_{{p_1}}} = 4A{e^{2t}}$
Уврштавајући у полазну једначину
$4A{e^{2t}} + 2A{e^{2t}} - 2A{e^{2t}} = 2{e^{2t}}$
$4A{e^{2t}} = 2{e^{2t}} {\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}A = \frac{1}{2}$
${y_{{p_1}}} = \frac{1}{2}{e^{2t}}$
Коначно, када вратимо смену добијамо
${y_{{p_1}}} = \frac{1}{2}{x^2}$
${f_2}\left( x \right) = 3t + 2$
$\left. \begin{gathered}
a = 0 \hfill \\
b = 0 \hfill \\
\end{gathered} \right\} \Rightarrow {\text{ }}a \pm bi = 0 {\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}r = 0$
$\left. \begin{gathered}
n = 1 \hfill \\
m = 0 \hfill \\
\end{gathered} \right\} \Rightarrow {\text{ }} {\text{ }}N = 1$
${y_{{p_2}}} = {t^0}{e^{0t}}\left( {\left( {At + B} \right)\cos 0t + \left( {Ct + D} \right)\sin 0t} \right)$
${y_{{p_2}}} = At + B$
${{\dot y}_{{p_2}}} = A$
${{\ddot y}_{{p_2}}} = 0$
$0 + A - 2\left( {At + B} \right) = 3t + 2$
$A = - \frac{3}{2} {\text{ }}{\text{ }} \wedge {\text{ }} {\text{ }} B = - \frac{7}{4}$
${y_{{p_2}}} = - \frac{3}{2}t - \frac{7}{4}$
${y_{{p_2}}} = - \frac{3}{2}\ln x - \frac{7}{4}$
Коначно решење наше полазне једначине је
\[Y = {y_h} + {y_{{p_1}}} + {y_{{p_2}}}\]
\[Y = \frac{{{C_1}}}{{{x^2}}} + {C_2}x + \frac{1}{2}{x^2} + - \frac{3}{2}\ln x - \frac{7}{4}\]
2. Реши диференцијалну једначину
${\left( {2 + 3x} \right)^2}y + 3\left( {2 + 3x} \right)y' - 4y = {\root 3 \of {2 + 3x} ^2} + 3\ln \left( {2 + 3x} \right)$.
$\left( {2 + 3x} \right) = {e^t}{\text{ }}{\text{ }} {\text{ }} \Rightarrow {\text{ }}{\text{ }} {\text{ }} t = \ln \left( {2 + 3x} \right)$
$3dx = {e^t}dt$
${{dt} \over {dx}} = {3 \over {{e^t}}} = 3{e^{ - t}}$
$y' = \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} =3 \dot y \cdot {e^{ - t}}$
$y'' = \frac{{dy'}}{{dx}} = \frac{{dy'}}{{dt}} \cdot \frac{{dt}}{{dx}} = \frac{d}{{dt}}\left( {3 \dot y \cdot {e^{ - t}}} \right) \cdot 3{e^{ - t}} = $
$ {\text{ }} {\text{ }}{\text{ }}{\text{ }}{\text{ }} = \left( {3 \ddot y{e^{ - t}} +3 \dot y\left( { - {e^{ - t}}} \right)} \right) \cdot {e^{ - t}} =9{e^{ - 2t}} \cdot \left( {\ddot y - \dot y} \right)$
Уврштавајући у полазну једначину добијамо
${e^{2t}} \cdot 9{e^{ - 2t}}\left( {\ddot y - \dot y} \right) + 3{e^t} \cdot 3{e^{ - t}}\dot y - 4y = \root 3 \of {{e^{2t}}} + 3\ln {e^t}$
$9\ddot y - 9\dot y + 9\dot y - 4y = {e^{{2 \over 3}t}} + 3t$
$9\ddot y - 4y = {e^{{2 \over 3}t}} + 3t$
$9\ddot y - 4y = 0$
$9{k^2} - 4 = 0$
${k_{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} = \pm {2 \over 3}$
${y_h} = {C_1}{e^{{2 \over 3}t}} + {C_2}{e^{ - {2 \over 3}t}}$
${f_1}\left( x \right) = {e^{{2 \over 3}t}}$
$\left. \begin{gathered}
a = {2 \over 3} \hfill \\
b = 0 \hfill \\
\end{gathered} \right\} \Rightarrow {\text{ }}a \pm bi = {2 \over 3}{\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}r = 1$
$\left. \begin{gathered}
n = 0 \hfill \\
m = 0 \hfill \\
\end{gathered} \right\} \Rightarrow {\text{ }} {\text{ }}N = 0$
${y_{{p_1}}} = {t^1}{e^{{2 \over 3}t}}\left( {A\cos 0t + B\sin 0t} \right)$
${y_{{p_1}}} = At{e^{{2 \over 3}t}}$
${{\dot y}_{{p_1}}} = A{e^{{2 \over 3}t}} + At{2 \over 3}{e^{{2 \over 3}t}} = {e^{{2 \over 3}t}}\left( {A + {2 \over 3}At} \right)$
${{\ddot y}_{{p_1}}} = {2 \over 3}{e^{{2 \over 3}t}}\left( {A + {2 \over 3}At} \right) + {e^{{2 \over 3}t}}{2 \over 3}A = {e^{{2 \over 3}t}}\left( {{4 \over 3}A + {4 \over 9}At} \right)$
$9{e^{{2 \over 3}t}}\left( {{4 \over 3}A + {4 \over 9}At} \right) - 4At{e^{{2 \over 3}t}} = {e^{{2 \over 3}t}}$
$A = {1 \over {12}}$
${y_{{p_1}}} = {1 \over {12}}t{e^{{2 \over 3}t}}$
${f_2}\left( x \right) = 3t$
$\left. \begin{gathered}
a = 0 \hfill \\
b = 0 \hfill \\
\end{gathered} \right\} \Rightarrow {\text{ }}a \pm bi = 0 {\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}r = 0$
$\left. \begin{gathered}
n = 1 \hfill \\
m = 0 \hfill \\
\end{gathered} \right\} \Rightarrow {\text{ }} {\text{ }}N = 1$
${y_{{p_2}}} = {t^0}{e^{0t}}\left( {\left( {At + B} \right)\cos 0t + \left( {Ct + D} \right)\sin 0t} \right)$
${y_{{p_2}}} = At + B$
${{\dot y}_{{p_2}}} = A$
${{\ddot y}_{{p_2}}} = 0$
$9 \cdot 0 - 4\left( {At + B} \right) = 3t$
$A = - \frac{3}{4} {\text{ }}{\text{ }} \wedge {\text{ }} {\text{ }} B = 0$
${y_{{p_1}}} = - {3 \over 4}t$
\[Y = {y_h} + {y_{{p_1}}} + {y_{{p_2}}}\]
$$Y = {C_1}{e^{{2 \over 3}t}} + {C_2}{e^{ - {2 \over 3}t}} + {1 \over {12}}t{e^{{2 \over 3}t}} - {3 \over 4}t$$
$$Y = {C_1}{\root 3 \of {2 + 3x} ^2} + {{{C_2}} \over {{{\root 3 \of {2 + 3x} }^2}}} + {{{{\root 3 \of {2 + 3x} }^2}} \over {12}}\ln \left( {2 + 3x} \right) - {3 \over 4}\ln \left( {2 + 3x} \right)$$