P d x + Q d y = 0 ако је ∂ P ∂ y = ∂ Q ∂ x тада дату једначину можемо записати у облику d F ( x , y ) = 0 односно ∂ F ∂ x ⏟ P d x + ∂ F ∂ y ⏟ Q d y = 0 где је F ( x , y ) = C \begin{array}{l} {\text{ }} {\text{ }} Pdx + Qdy = 0 \\ {\text{ако је }} {\text{ }}\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}} \\ {\text{тада дату једначину можемо записати у облику}} \\ {\text{ }} {\text{ }}{\text{ }} {\text{ }}{\text{ }} {\text{ }} dF\left( {x,y} \right) = 0 \\ односно{\text{ }} {\text{ }} \underbrace {\frac{{\partial F}}{{\partial x}}}_Pdx + \underbrace {\frac{{\partial F}}{{\partial y}}}_Qdy = 0 \\ {\text{где је }} {\text{ }} F\left( {x,y} \right) = C \\ \end{array} P d x + Q d y = 0 ако је ∂ y ∂ P = ∂ x ∂ Q тада дату једначину можемо записати у облику d F ( x , y ) = 0 односно P ∂ x ∂ F d x + Q ∂ y ∂ F d y = 0 где је F ( x , y ) = C
Задаци 1. Реши диференцијалну једначину x 2 y " + 2 x y ′ − 2 y = 2 x 2 + 3 ln x + 2 {x^2}y" + 2xy' - 2y = 2{x^2} + 3\ln x + 2 x 2 y " + 2 x y ′ − 2 y = 2 x 2 + 3 ln x + 2
Препознајемо да је дата диференцијална једначина Ојлерова, па уводимо смену: x = e t ⇒ t = ln x x = {e^t}{\text{ }}{\text{ }} {\text{ }} \Rightarrow {\text{ }}{\text{ }} {\text{ }} t = \ln x x = e t ⇒ t = ln x
d x = e t d t dx = {e^t}dt d x = e t d t
d t d x = e − t \frac{{dt}}{{dx}} = {e^{ - t}} d x d t = e − t
y ′ = d y d x = d y d t ⋅ d t d x = y ˙ ⋅ e − t y' = \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} = \dot y \cdot {e^{ - t}} y ′ = d x d y = d t d y ⋅ d x d t = y ˙ ⋅ e − t
y " = d y ′ d x = d y ′ d t ⋅ d t d x = d d t ( y ˙ ⋅ e − t ) ⋅ 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}} = y " = d x d y ′ = d t d y ′ ⋅ d x d t = d t d ( y ˙ ⋅ e − t ) ⋅ e − t =
= ( y ¨ e − t + y ˙ ( − e − t ) ) ⋅ e − t = ( y ¨ − y ˙ ) ⋅ e − 2 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}} = ( y ¨ e − t + y ˙ ( − e − t ) ) ⋅ e − t = ( y ¨ − y ˙ ) ⋅ e − 2 t
Уврштавајући у полазну једначину добијамо ( y ¨ e − t + y ˙ ( − e − t ) ) ⋅ e − t = ( y ¨ − y ˙ ) ⋅ e − 2 t \left( {\ddot y{e^{ - t}} + \dot y\left( { - {e^{ - t}}} \right)} \right) \cdot {e^{ - t}} = \left( {\ddot y - \dot y} \right) \cdot {e^{ - 2t}} ( y ¨ e − t + y ˙ ( − e − t ) ) ⋅ e − t = ( y ¨ − y ˙ ) ⋅ e − 2 t
e 2 t ( y ¨ − y ˙ ) ⋅ e − 2 t + 2 e t y ˙ ⋅ e − t − 2 y = 2 e 2 t + 3 t + 2 {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 e 2 t ( y ¨ − y ˙ ) ⋅ e − 2 t + 2 e t y ˙ ⋅ e − t − 2 y = 2 e 2 t + 3 t + 2
y ¨ − y ˙ + 2 y ˙ − 2 y = 2 e 2 t + 3 t + 2 \ddot y - \dot y + 2\dot y - 2y = 2{e^{2t}} + 3t + 2 y ¨ − y ˙ + 2 y ˙ − 2 y = 2 e 2 t + 3 t + 2
y ¨ + y ˙ − 2 y = 2 e 2 t + 3 t + 2 \ddot y + \dot y - 2y = 2{e^{2t}} + 3t + 2 y ¨ + y ˙ − 2 y = 2 e 2 t + 3 t + 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^2} + k - 2 = 0 k 2 + k − 2 = 0
k 1 = − 2 ∧ k 2 = 1 {k_1} = - 2{\text{ }}{\text{ }} \wedge {\text{ }}{\text{ }} {k_2} = 1 k 1 = − 2 ∧ k 2 = 1
y h = C 1 e − 2 t + C 2 e t {y_h} = {C_1}{e^{ - 2t}} + {C_2}{e^t} y h = C 1 e − 2 t + C 2 e t
Враћајући смену t = ln x t = \ln x t = ln x y h = C 1 x 2 + C 2 x {y_h} = \frac{{{C_1}}}{{{x^2}}} + {C_2}x y h = x 2 C 1 + C 2 x
f 1 ( x ) = 2 e 2 t {f_1}\left( x \right) = 2{e^{2t}} f 1 ( x ) = 2 e 2 t
a = 2 b = 0 } ⇒ a ± b i = 2 ⇒ r = 0 \left. \begin{aligned}&a = 2 \\ &b = 0\end{aligned} \right\} \Rightarrow {\text{ }}a \pm bi = 2{\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}r = 0 a = 2 b = 0 } ⇒ a ± bi = 2 ⇒ r = 0
n = 0 m = 0 } ⇒ N = 0 \left. \begin{aligned}&n = 0 \\ &m = 0\end{aligned} \right\} \Rightarrow {\text{ }} {\text{ }}N = 0 n = 0 m = 0 } ⇒ N = 0
y p 1 = t 0 e 2 t ( A cos 0 t + B sin 0 t ) {y_{{p_1}}} = {t^0}{e^{2t}}\left( {A\cos 0t + B\sin 0t} \right) y p 1 = t 0 e 2 t ( A cos 0 t + B sin 0 t )
y p 1 = A e 2 t {y_{{p_1}}} = A{e^{2t}} y p 1 = A e 2 t
y ˙ p 1 = 2 A e 2 t {{\dot y}_{{p_1}}} = 2A{e^{2t}} y ˙ p 1 = 2 A e 2 t
y ¨ p 1 = 4 A e 2 t {{\ddot y}_{{p_1}}} = 4A{e^{2t}} y ¨ p 1 = 4 A e 2 t
Уврштавајући у полазну једначину 4 A e 2 t + 2 A e 2 t − 2 A e 2 t = 2 e 2 t 4A{e^{2t}} + 2A{e^{2t}} - 2A{e^{2t}} = 2{e^{2t}} 4 A e 2 t + 2 A e 2 t − 2 A e 2 t = 2 e 2 t
4 A e 2 t = 2 e 2 t ⇒ A = 1 2 4A{e^{2t}} = 2{e^{2t}} {\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}A = \frac{1}{2} 4 A e 2 t = 2 e 2 t ⇒ A = 2 1
y p 1 = 1 2 e 2 t {y_{{p_1}}} = \frac{1}{2}{e^{2t}} y p 1 = 2 1 e 2 t
Коначно, када вратимо смену добијамо y p 1 = 1 2 x 2 {y_{{p_1}}} = \frac{1}{2}{x^2} y p 1 = 2 1 x 2
f 2 ( x ) = 3 t + 2 {f_2}\left( x \right) = 3t + 2 f 2 ( x ) = 3 t + 2
a = 0 b = 0 } ⇒ a ± b i = 0 ⇒ r = 0 \left. \begin{aligned}&a = 0 \\ &b = 0\end{aligned} \right\} \Rightarrow {\text{ }}a \pm bi = 0 {\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}r = 0 a = 0 b = 0 } ⇒ a ± bi = 0 ⇒ r = 0
n = 1 m = 0 } ⇒ N = 1 \left. \begin{aligned}&n = 1 \\ &m = 0\end{aligned} \right\} \Rightarrow {\text{ }} {\text{ }}N = 1 n = 1 m = 0 } ⇒ N = 1
y p 2 = t 0 e 0 t ( ( A t + B ) cos 0 t + ( C t + D ) sin 0 t ) {y_{{p_2}}} = {t^0}{e^{0t}}\left( {\left( {At + B} \right)\cos 0t + \left( {Ct + D} \right)\sin 0t} \right) y p 2 = t 0 e 0 t ( ( A t + B ) cos 0 t + ( C t + D ) sin 0 t )
y p 2 = A t + B {y_{{p_2}}} = At + B y p 2 = A t + B
y ˙ p 2 = A {{\dot y}_{{p_2}}} = A y ˙ p 2 = A
y ¨ p 2 = 0 {{\ddot y}_{{p_2}}} = 0 y ¨ p 2 = 0
0 + A − 2 ( A t + B ) = 3 t + 2 0 + A - 2\left( {At + B} \right) = 3t + 2 0 + A − 2 ( A t + B ) = 3 t + 2
A = − 3 2 ∧ B = − 7 4 A = - \frac{3}{2} {\text{ }}{\text{ }} \wedge {\text{ }} {\text{ }} B = - \frac{7}{4} A = − 2 3 ∧ B = − 4 7
y p 2 = − 3 2 t − 7 4 {y_{{p_2}}} = - \frac{3}{2}t - \frac{7}{4} y p 2 = − 2 3 t − 4 7
y p 2 = − 3 2 ln x − 7 4 {y_{{p_2}}} = - \frac{3}{2}\ln x - \frac{7}{4} y p 2 = − 2 3 ln x − 4 7
Коначно решење наше полазне једначине је Y = y h + y p 1 + y p 2 Y = {y_h} + {y_{{p_1}}} + {y_{{p_2}}} Y = y h + y p 1 + y p 2
Y = C 1 x 2 + C 2 x + 1 2 x 2 + − 3 2 ln x − 7 4 Y = \frac{{{C_1}}}{{{x^2}}} + {C_2}x + \frac{1}{2}{x^2} + - \frac{3}{2}\ln x - \frac{7}{4} Y = x 2 C 1 + C 2 x + 2 1 x 2 + − 2 3 ln x − 4 7
2. Реши диференцијалну једначину
( 2 + 3 x ) 2 y + 3 ( 2 + 3 x ) y ′ − 4 y = 2 + 3 x 3 2 + 3 ln ( 2 + 3 x ) {\left( {2 + 3x} \right)^2}y + 3\left( {2 + 3x} \right)y' - 4y = {\sqrt[3]{2 + 3x} ^2} + 3\ln \left( {2 + 3x} \right) ( 2 + 3 x ) 2 y + 3 ( 2 + 3 x ) y ′ − 4 y = 3 2 + 3 x 2 + 3 ln ( 2 + 3 x )
( 2 + 3 x ) = e t ⇒ t = ln ( 2 + 3 x ) \left( {2 + 3x} \right) = {e^t}{\text{ }}{\text{ }} {\text{ }} \Rightarrow {\text{ }}{\text{ }} {\text{ }} t = \ln \left( {2 + 3x} \right) ( 2 + 3 x ) = e t ⇒ t = ln ( 2 + 3 x )
3 d x = e t d t 3dx = {e^t}dt 3 d x = e t d t
d t d x = 3 e t = 3 e − t {{dt} \over {dx}} = {3 \over {{e^t}}} = 3{e^{ - t}} d x d t = e t 3 = 3 e − t
y ′ = d y d x = d y d t ⋅ d t d x = 3 y ˙ ⋅ e − t y' = \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \cdot \frac{{dt}}{{dx}} =3 \dot y \cdot {e^{ - t}} y ′ = d x d y = d t d y ⋅ d x d t = 3 y ˙ ⋅ e − t
y " = d y ′ d x = d y ′ d t ⋅ d t d x = d d t ( 3 y ˙ ⋅ e − t ) ⋅ 3 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}} = y " = d x d y ′ = d t d y ′ ⋅ d x d t = d t d ( 3 y ˙ ⋅ e − t ) ⋅ 3 e − t =
= ( 3 y ¨ e − t + 3 y ˙ ( − e − t ) ) ⋅ e − t = 9 e − 2 t ⋅ ( y ¨ − y ˙ ) {\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) = ( 3 y ¨ e − t + 3 y ˙ ( − e − t ) ) ⋅ e − t = 9 e − 2 t ⋅ ( y ¨ − y ˙ )
Уврштавајући у полазну једначину добијамо e 2 t ⋅ 9 e − 2 t ( y ¨ − y ˙ ) + 3 e t ⋅ 3 e − t y ˙ − 4 y = e 2 t 3 + 3 ln e t {e^{2t}} \cdot 9{e^{ - 2t}}\left( {\ddot y - \dot y} \right) + 3{e^t} \cdot 3{e^{ - t}}\dot y - 4y = \sqrt[3]{{e^{2t}}} + 3\ln {e^t} e 2 t ⋅ 9 e − 2 t ( y ¨ − y ˙ ) + 3 e t ⋅ 3 e − t y ˙ − 4 y = 3 e 2 t + 3 ln e t
9 y ¨ − 9 y ˙ + 9 y ˙ − 4 y = e 2 3 t + 3 t 9\ddot y - 9\dot y + 9\dot y - 4y = {e^{{2 \over 3}t}} + 3t 9 y ¨ − 9 y ˙ + 9 y ˙ − 4 y = e 3 2 t + 3 t
9 y ¨ − 4 y = e 2 3 t + 3 t 9\ddot y - 4y = {e^{{2 \over 3}t}} + 3t 9 y ¨ − 4 y = e 3 2 t + 3 t
9 y ¨ − 4 y = 0 9\ddot y - 4y = 0 9 y ¨ − 4 y = 0
9 k 2 − 4 = 0 9{k^2} - 4 = 0 9 k 2 − 4 = 0
k 1 2 = ± 2 3 {k_{{\tfrac{1}{2}}}} = \pm {2 \over 3} k 2 1 = ± 3 2
y h = C 1 e 2 3 t + C 2 e − 2 3 t {y_h} = {C_1}{e^{{2 \over 3}t}} + {C_2}{e^{ - {2 \over 3}t}} y h = C 1 e 3 2 t + C 2 e − 3 2 t
f 1 ( x ) = e 2 3 t {f_1}\left( x \right) = {e^{{2 \over 3}t}} f 1 ( x ) = e 3 2 t
a = 2 3 b = 0 } ⇒ a ± b i = 2 3 ⇒ r = 1 \left. \begin{aligned}&a = {2 \over 3} \\ &b = 0\end{aligned} \right\} \Rightarrow {\text{ }}a \pm bi = {2 \over 3}{\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}r = 1 a = 3 2 b = 0 ⎭ ⎬ ⎫ ⇒ a ± bi = 3 2 ⇒ r = 1
n = 0 m = 0 } ⇒ N = 0 \left. \begin{aligned}&n = 0 \\ &m = 0\end{aligned} \right\} \Rightarrow {\text{ }} {\text{ }}N = 0 n = 0 m = 0 } ⇒ N = 0
y p 1 = t 1 e 2 3 t ( A cos 0 t + B sin 0 t ) {y_{{p_1}}} = {t^1}{e^{{2 \over 3}t}}\left( {A\cos 0t + B\sin 0t} \right) y p 1 = t 1 e 3 2 t ( A cos 0 t + B sin 0 t )
y p 1 = A t e 2 3 t {y_{{p_1}}} = At{e^{{2 \over 3}t}} y p 1 = A t e 3 2 t
y ˙ p 1 = A e 2 3 t + A t 2 3 e 2 3 t = e 2 3 t ( A + 2 3 A 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) y ˙ p 1 = A e 3 2 t + A t 3 2 e 3 2 t = e 3 2 t ( A + 3 2 A t )
y ¨ p 1 = 2 3 e 2 3 t ( A + 2 3 A t ) + e 2 3 t 2 3 A = e 2 3 t ( 4 3 A + 4 9 A t ) {{\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) y ¨ p 1 = 3 2 e 3 2 t ( A + 3 2 A t ) + e 3 2 t 3 2 A = e 3 2 t ( 3 4 A + 9 4 A t )
9 e 2 3 t ( 4 3 A + 4 9 A t ) − 4 A t e 2 3 t = e 2 3 t 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}} 9 e 3 2 t ( 3 4 A + 9 4 A t ) − 4 A t e 3 2 t = e 3 2 t
A = 1 12 A = {1 \over {12}} A = 12 1
y p 1 = 1 12 t e 2 3 t {y_{{p_1}}} = {1 \over {12}}t{e^{{2 \over 3}t}} y p 1 = 12 1 t e 3 2 t
f 2 ( x ) = 3 t {f_2}\left( x \right) = 3t f 2 ( x ) = 3 t
a = 0 b = 0 } ⇒ a ± b i = 0 ⇒ r = 0 \left. \begin{aligned}&a = 0 \\ &b = 0\end{aligned} \right\} \Rightarrow {\text{ }}a \pm bi = 0 {\text{ }}{\text{ }} \Rightarrow {\text{ }} {\text{ }}r = 0 a = 0 b = 0 } ⇒ a ± bi = 0 ⇒ r = 0
n = 1 m = 0 } ⇒ N = 1 \left. \begin{aligned}&n = 1 \\ &m = 0\end{aligned} \right\} \Rightarrow {\text{ }} {\text{ }}N = 1 n = 1 m = 0 } ⇒ N = 1
y p 2 = t 0 e 0 t ( ( A t + B ) cos 0 t + ( C t + D ) sin 0 t ) {y_{{p_2}}} = {t^0}{e^{0t}}\left( {\left( {At + B} \right)\cos 0t + \left( {Ct + D} \right)\sin 0t} \right) y p 2 = t 0 e 0 t ( ( A t + B ) cos 0 t + ( C t + D ) sin 0 t )
y p 2 = A t + B {y_{{p_2}}} = At + B y p 2 = A t + B
y ˙ p 2 = A {{\dot y}_{{p_2}}} = A y ˙ p 2 = A
y ¨ p 2 = 0 {{\ddot y}_{{p_2}}} = 0 y ¨ p 2 = 0
9 ⋅ 0 − 4 ( A t + B ) = 3 t 9 \cdot 0 - 4\left( {At + B} \right) = 3t 9 ⋅ 0 − 4 ( A t + B ) = 3 t
A = − 3 4 ∧ B = 0 A = - \frac{3}{4} {\text{ }}{\text{ }} \wedge {\text{ }} {\text{ }} B = 0 A = − 4 3 ∧ B = 0
y p 1 = − 3 4 t {y_{{p_1}}} = - {3 \over 4}t y p 1 = − 4 3 t
Y = y h + y p 1 + y p 2 Y = {y_h} + {y_{{p_1}}} + {y_{{p_2}}} Y = y h + y p 1 + y p 2
Y = C 1 e 2 3 t + C 2 e − 2 3 t + 1 12 t e 2 3 t − 3 4 t 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 e 3 2 t + C 2 e − 3 2 t + 12 1 t e 3 2 t − 4 3 t
Y = C 1 2 + 3 x 3 2 + C 2 2 + 3 x 3 2 + 2 + 3 x 3 2 12 ln ( 2 + 3 x ) − 3 4 ln ( 2 + 3 x ) Y = {C_1}{\sqrt[3]{2 + 3x} ^2} + {{{C_2}} \over {{{\sqrt[3]{2 + 3x} }^2}}} + {{{{\sqrt[3]{2 + 3x} }^2}} \over {12}}\ln \left( {2 + 3x} \right) - {3 \over 4}\ln \left( {2 + 3x} \right) Y = C 1 3 2 + 3 x 2 + 3 2 + 3 x 2 C 2 + 12 3 2 + 3 x 2 ln ( 2 + 3 x ) − 4 3 ln ( 2 + 3 x )