ANSWER on Question #80273 – Math – Differential Equations
QUESTION
Solve the PDE:
1)
( D 2 + D − 1 ) z = 4 e x + y cos ( x + y ) (D^2 + D - 1)z = 4e^{x+y} \cos(x + y) ( D 2 + D − 1 ) z = 4 e x + y cos ( x + y )
2)
( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) z = 4 + 3 x + 6 y (D + D' - 1)(D + 2D' - 3)z = 4 + 3x + 6y ( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) z = 4 + 3 x + 6 y SOLUTION
Before solving the task, let us recall some theoretical facts.
Let the given differential equation be
F ( D , D ′ ) = f ( x , y ) . F(D, D') = f(x, y). F ( D , D ′ ) = f ( x , y ) .
Factorize F ( D , D ′ ) F(D, D') F ( D , D ′ )
**Rule I.** Corresponding to each non-repeated factor ( b D − a D ′ − c ) (bD - aD' - c) ( b D − a D ′ − c )
e ( c x b ) φ ( b y + a x ) , if b ≠ 0 e^{\left(\frac{cx}{b}\right)} \varphi(by + ax), \quad \text{if } b \neq 0 e ( b c x ) φ ( b y + a x ) , if b = 0
We now have three particular cases of Rule I:
**Rule IA.** Take c = 0 c = 0 c = 0 ( b D − a D ′ ) (bD - aD') ( b D − a D ′ )
φ ( b y + a x ) , if b ≠ 0. \varphi(by + ax), \quad \text{if } b \neq 0. φ ( b y + a x ) , if b = 0.
**Rule IB.** Take a = 0 a = 0 a = 0 ( b D − c ) (bD - c) ( b D − c )
e ( c x b ) φ ( b y ) , if b ≠ 0. e^{\left(\frac{cx}{b}\right)} \varphi(by), \quad \text{if } b \neq 0. e ( b c x ) φ ( b y ) , if b = 0.
**Rule IC.** Take a = c = 0 a = c = 0 a = c = 0 b = 1 b = 1 b = 1 ( 1 ⋅ D ) (1 \cdot D) ( 1 ⋅ D )
φ ( y ) . \varphi(y). φ ( y ) .
Rule II. When
f ( x , y ) = V ( x , y ) e a x + b y f(x, y) = V(x, y) e^{a x + b y} f ( x , y ) = V ( x , y ) e a x + b y
Then,
P . I . = 1 F ( D , D ′ ) V e a x + b y = e a x + b y 1 F ( D + a , D ′ + b ) V ( x , y ) P.I. = \frac{1}{F(D, D')} V e^{a x + b y} = e^{a x + b y} \frac{1}{F(D + a, D' + b)} V(x, y) P . I . = F ( D , D ′ ) 1 V e a x + b y = e a x + b y F ( D + a , D ′ + b ) 1 V ( x , y )
In our case,
1)
( D 2 + D − 1 ) z = 4 e x + y cos ( x + y ) (D^2 + D - 1)z = 4 e^{x + y} \cos(x + y) ( D 2 + D − 1 ) z = 4 e x + y cos ( x + y ) ( D 2 + D − 1 ) z = 4 e x + y cos ( x + y ) → z ( x , y ) = C . F . + P . I . (D^2 + D - 1)z = 4 e^{x + y} \cos(x + y) \rightarrow z(x, y) = C.F. + P.I. ( D 2 + D − 1 ) z = 4 e x + y cos ( x + y ) → z ( x , y ) = C . F . + P . I .
O STEP: Factorize F ( D , D ′ ) F(D, D') F ( D , D ′ )
D 2 + D − 1 → 1 ⏟ a ⋅ x 2 + 1 ⏟ b ⋅ x − 1 ⏟ c = 0 → D = b 2 − 4 a c = ( 1 ) 2 − 4 ⋅ 1 ⋅ ( − 1 ) = 1 + 4 D^2 + D - 1 \rightarrow \underbrace{1}_{a} \cdot x^2 + \underbrace{1}_{b} \cdot x \underbrace{-1}_{c} = 0 \rightarrow \sqrt{D} = \sqrt{b^2 - 4ac} = \sqrt{(1)^2 - 4 \cdot 1 \cdot (-1)} = \sqrt{1 + 4} D 2 + D − 1 → a 1 ⋅ x 2 + b 1 ⋅ x c − 1 = 0 → D = b 2 − 4 a c = ( 1 ) 2 − 4 ⋅ 1 ⋅ ( − 1 ) = 1 + 4 D = 5 → [ m 1 = − b − D 2 a = − 1 − 5 2 ⋅ 1 = − 1 + 5 2 m 2 = − b + D 2 a = − 1 + 5 2 ⋅ 1 = − 1 − 5 2 ] \sqrt{D} = \sqrt{5} \rightarrow \begin{bmatrix}
m_1 = \dfrac{-b - \sqrt{D}}{2a} = \dfrac{-1 - \sqrt{5}}{2 \cdot 1} = -\dfrac{1 + \sqrt{5}}{2} \\
m_2 = \dfrac{-b + \sqrt{D}}{2a} = \dfrac{-1 + \sqrt{5}}{2 \cdot 1} = -\dfrac{1 - \sqrt{5}}{2}
\end{bmatrix} D = 5 → ⎣ ⎡ m 1 = 2 a − b − D = 2 ⋅ 1 − 1 − 5 = − 2 1 + 5 m 2 = 2 a − b + D = 2 ⋅ 1 − 1 + 5 = − 2 1 − 5 ⎦ ⎤
Conclusion,
m 2 + m − 1 = ( m − m 1 ) ( m − m 2 ) = ( m + 1 + 5 2 ) ( m + 1 − 5 2 ) → m^2 + m - 1 = (m - m_1)(m - m_2) = \left(m + \dfrac{1 + \sqrt{5}}{2}\right) \left(m + \dfrac{1 - \sqrt{5}}{2}\right) \rightarrow m 2 + m − 1 = ( m − m 1 ) ( m − m 2 ) = ( m + 2 1 + 5 ) ( m + 2 1 − 5 ) → D 2 + D − 1 = ( D + 1 + 5 2 ) ( D + 1 − 5 2 ) D^2 + D - 1 = \left(D + \dfrac{1 + \sqrt{5}}{2}\right) \left(D + \dfrac{1 - \sqrt{5}}{2}\right) D 2 + D − 1 = ( D + 2 1 + 5 ) ( D + 2 1 − 5 )
Then,
D 2 + D − 1 z = 4 e x + y cos ( x + y ) → ( D + 1 + 5 2 ) ( D + 1 − 5 2 ) z = 4 e x + y cos ( x + y ) D^2 + D - 1 z = 4 e^{x + y} \cos(x + y) \rightarrow \left(D + \dfrac{1 + \sqrt{5}}{2}\right) \left(D + \dfrac{1 - \sqrt{5}}{2}\right) z = 4 e^{x + y} \cos(x + y) D 2 + D − 1 z = 4 e x + y cos ( x + y ) → ( D + 2 1 + 5 ) ( D + 2 1 − 5 ) z = 4 e x + y cos ( x + y )
1 STEP: Let find C.F.
{ ( D + 1 + 5 2 ) z ( b D − a D ′ − c ) z → { b = 1 a = 0 c = − 1 + 5 2 → ( C . F . ) 1 = e ( − 1 + 5 2 x 1 ) ⋅ φ 1 ( 1 ⋅ y + ( 0 ) ⋅ x ) → \left\{ \begin{array}{l} \left(D + \frac {1 + \sqrt {5}}{2}\right) z \\ (b D - a D ^ {\prime} - c) z \end{array} \right. \to \left\{ \begin{array}{c} b = 1 \\ a = 0 \\ c = - \frac {1 + \sqrt {5}}{2} \end{array} \right. \to (C. F.) _ {1} = e ^ {\left(\frac {- \frac {1 + \sqrt {5}}{2} x}{1}\right)} \cdot \varphi_ {1} (1 \cdot y + (0) \cdot x) \to { ( D + 2 1 + 5 ) z ( b D − a D ′ − c ) z → ⎩ ⎨ ⎧ b = 1 a = 0 c = − 2 1 + 5 → ( C . F . ) 1 = e ( 1 − 2 1 + 5 x ) ⋅ φ 1 ( 1 ⋅ y + ( 0 ) ⋅ x ) → C . F . ) 1 = e − 1 + 5 2 x ⋅ φ 1 ( y ) , where φ 1 is arbitrary function C. F.) _ {1} = e ^ {- \frac {1 + \sqrt {5}}{2} x} \cdot \varphi_ {1} (y), \text{where } \varphi_ {1} \text{ is arbitrary function} C . F . ) 1 = e − 2 1 + 5 x ⋅ φ 1 ( y ) , where φ 1 is arbitrary function { ( D + 1 − 5 2 ) z ( b D − a D ′ − c ) z → { b = 1 a = 0 c = − 1 − 5 2 → ( C . F . ) 2 = e ( − 1 − 5 2 x 1 ) ⋅ φ 2 ( 1 ⋅ y + ( 0 ) ⋅ x ) → \left\{ \begin{array}{l} \left(D + \frac {1 - \sqrt {5}}{2}\right) z \\ (b D - a D ^ {\prime} - c) z \end{array} \right. \to \left\{ \begin{array}{c} b = 1 \\ a = 0 \\ c = - \frac {1 - \sqrt {5}}{2} \end{array} \right. \to (C. F.) _ {2} = e ^ {\left(\frac {- \frac {1 - \sqrt {5}}{2} x}{1}\right)} \cdot \varphi_ {2} (1 \cdot y + (0) \cdot x) \to { ( D + 2 1 − 5 ) z ( b D − a D ′ − c ) z → ⎩ ⎨ ⎧ b = 1 a = 0 c = − 2 1 − 5 → ( C . F . ) 2 = e ( 1 − 2 1 − 5 x ) ⋅ φ 2 ( 1 ⋅ y + ( 0 ) ⋅ x ) → C . F . ) 2 = e − 1 − 5 2 x ⋅ φ 2 ( y ) , where φ 2 is arbitrary function C. F.) _ {2} = e ^ {- \frac {1 - \sqrt {5}}{2} x} \cdot \varphi_ {2} (y), \text{where } \varphi_ {2} \text{ is arbitrary function} C . F . ) 2 = e − 2 1 − 5 x ⋅ φ 2 ( y ) , where φ 2 is arbitrary function
Then,
C . F . = ( C . F . ) 1 + ( C . F . ) 2 → C . F . = e − 1 + 5 2 x ⋅ φ 1 ( y ) + e − 1 − 5 2 x ⋅ φ 2 ( y ) C. F. = (C. F.) _ {1} + (C. F.) _ {2} \rightarrow C. F. = e ^ {- \frac {1 + \sqrt {5}}{2} x} \cdot \varphi_ {1} (y) + e ^ {- \frac {1 - \sqrt {5}}{2} x} \cdot \varphi_ {2} (y) C . F . = ( C . F . ) 1 + ( C . F . ) 2 → C . F . = e − 2 1 + 5 x ⋅ φ 1 ( y ) + e − 2 1 − 5 x ⋅ φ 2 ( y )
2 STEP: Let find P.I.
P . I . = 1 D 2 + D − 1 4 e a 1 ⋅ x + b 1 ⋅ y cos ( x + y ) = 4 e x + y 1 ( D + 1 ) 2 + ( D + 1 ) − 1 cos ( x + y ) = = 4 e x + y 1 D 2 + 2 D + 1 + D + 1 − 1 cos ( x + y ) = 4 e x + y 1 D 2 + 3 D + 1 cos ( x + y ) = = 4 e x + y 1 ( − 1 ) 2 + 3 D + 1 cos ( x + y ) = 4 e x + y 1 1 + 3 D + 1 cos ( x + y ) = = 4 e x + y 1 3 D + 2 cos ( x + y ) = 4 e x + y ( 3 D − 2 ) ( 3 D + 2 ) ( 3 D − 2 ) cos ( x + y ) = = 4 e x + y ( 3 D − 2 ) 9 D 2 − 4 cos ( x + y ) = 4 e x + y ( 3 D − 2 ) 9 ⋅ ( − 1 ) 2 − 4 cos ( x + y ) = = 4 e x + y ( 3 D − 2 ) 9 ⋅ 1 − 4 cos ( x + y ) = 4 e x + y ( 3 D − 2 ) 5 cos ( x + y ) = = 4 5 e x + y ( 3 D − 2 ) cos ( x + y ) = 4 5 e x + y ( 3 ∂ ∂ x − 2 ) cos ( x + y ) = = 4 5 e x + y ( 3 ∂ ∂ x ( cos ( x + y ) ) − 2 ⋅ cos ( x + y ) ) = \begin{array}{l} P. I. = \frac {1}{D ^ {2} + D - 1} 4 e ^ {\frac {a}{1} \cdot x + \frac {b}{1} \cdot y} \cos (x + y) = 4 e ^ {x + y} \frac {1}{(D + 1) ^ {2} + (D + 1) - 1} \cos (x + y) = \\ = 4 e ^ {x + y} \frac {1}{D ^ {2} + 2 D + 1 + D + 1 - 1} \cos (x + y) = 4 e ^ {x + y} \frac {1}{D ^ {2} + 3 D + 1} \cos (x + y) = \\ = 4 e ^ {x + y} \frac {1}{(- 1) ^ {2} + 3 D + 1} \cos (x + y) = 4 e ^ {x + y} \frac {1}{1 + 3 D + 1} \cos (x + y) = \\ = 4 e ^ {x + y} \frac {1}{3 D + 2} \cos (x + y) = 4 e ^ {x + y} \frac {(3 D - 2)}{(3 D + 2) (3 D - 2)} \cos (x + y) = \\ = 4 e ^ {x + y} \frac {(3 D - 2)}{9 D ^ {2} - 4} \cos (x + y) = 4 e ^ {x + y} \frac {(3 D - 2)}{9 \cdot (- 1) ^ {2} - 4} \cos (x + y) = = 4 e ^ {x + y} \frac {(3 D - 2)}{9 \cdot 1 - 4} \cos (x + y) \\ = 4 e ^ {x + y} \frac {(3 D - 2)}{5} \cos (x + y) = \\ = \frac {4}{5} e ^ {x + y} (3 D - 2) \cos (x + y) = \frac {4}{5} e ^ {x + y} \left(3 \frac {\partial}{\partial x} - 2\right) \cos (x + y) = \\ = \frac {4}{5} e ^ {x + y} \left(3 \frac {\partial}{\partial x} (\cos (x + y)) - 2 \cdot \cos (x + y)\right) = \\ \end{array} P . I . = D 2 + D − 1 1 4 e 1 a ⋅ x + 1 b ⋅ y cos ( x + y ) = 4 e x + y ( D + 1 ) 2 + ( D + 1 ) − 1 1 cos ( x + y ) = = 4 e x + y D 2 + 2 D + 1 + D + 1 − 1 1 cos ( x + y ) = 4 e x + y D 2 + 3 D + 1 1 cos ( x + y ) = = 4 e x + y ( − 1 ) 2 + 3 D + 1 1 cos ( x + y ) = 4 e x + y 1 + 3 D + 1 1 cos ( x + y ) = = 4 e x + y 3 D + 2 1 cos ( x + y ) = 4 e x + y ( 3 D + 2 ) ( 3 D − 2 ) ( 3 D − 2 ) cos ( x + y ) = = 4 e x + y 9 D 2 − 4 ( 3 D − 2 ) cos ( x + y ) = 4 e x + y 9 ⋅ ( − 1 ) 2 − 4 ( 3 D − 2 ) cos ( x + y ) == 4 e x + y 9 ⋅ 1 − 4 ( 3 D − 2 ) cos ( x + y ) = 4 e x + y 5 ( 3 D − 2 ) cos ( x + y ) = = 5 4 e x + y ( 3 D − 2 ) cos ( x + y ) = 5 4 e x + y ( 3 ∂ x ∂ − 2 ) cos ( x + y ) = = 5 4 e x + y ( 3 ∂ x ∂ ( cos ( x + y )) − 2 ⋅ cos ( x + y ) ) = = 4 5 e x + y ( 3 ( − sin ( x + y ) ) − 2 cos ( x + y ) ) = = − 4 5 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y ) ) \begin{array}{l}
= \frac {4}{5} e ^ {x + y} (3 (- \sin (x + y)) - 2 \cos (x + y)) = \\
= - \frac {4}{5} e ^ {x + y} (3 \sin (x + y) + 2 \cos (x + y))
\end{array} = 5 4 e x + y ( 3 ( − sin ( x + y )) − 2 cos ( x + y )) = = − 5 4 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y ))
Then,
P . I . = − 4 5 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y ) ) P.I. = - \frac {4}{5} e ^ {x + y} (3 \sin (x + y) + 2 \cos (x + y)) P . I . = − 5 4 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y ))
Conclusion,
z ( x , y ) = C . F . + P . I . = = e − 1 + 5 2 x ⋅ φ 1 ( y ) + e − 1 − 5 2 x ⋅ φ 2 ( y ) − 4 5 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y ) ) z ( x , y ) = e − 1 + 5 2 x ⋅ φ 1 ( y ) + e − 1 − 5 2 x ⋅ φ 2 ( y ) − 4 5 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y ) ) \begin{array}{l}
z (x, y) = C. F. + P. I. = \\
= e ^ {- \frac {1 + \sqrt {5}}{2} x} \cdot \varphi_ {1} (y) + e ^ {- \frac {1 - \sqrt {5}}{2} x} \cdot \varphi_ {2} (y) - \frac {4}{5} e ^ {x + y} (3 \sin (x + y) + 2 \cos (x + y)) \\
z (x, y) = e ^ {- \frac {1 + \sqrt {5}}{2} x} \cdot \varphi_ {1} (y) + e ^ {- \frac {1 - \sqrt {5}}{2} x} \cdot \varphi_ {2} (y) - \frac {4}{5} e ^ {x + y} (3 \sin (x + y) + 2 \cos (x + y))
\end{array} z ( x , y ) = C . F . + P . I . = = e − 2 1 + 5 x ⋅ φ 1 ( y ) + e − 2 1 − 5 x ⋅ φ 2 ( y ) − 5 4 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y )) z ( x , y ) = e − 2 1 + 5 x ⋅ φ 1 ( y ) + e − 2 1 − 5 x ⋅ φ 2 ( y ) − 5 4 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y )) { z ( x , y ) = e − 1 + 5 2 x ⋅ φ 1 ( y ) + e − 1 − 5 2 x ⋅ φ 2 ( y ) − 4 5 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y ) ) where φ 1 and φ 2 are arbitrary functions \left\{ \begin{array}{l}
z (x, y) = e ^ {- \frac {1 + \sqrt {5}}{2} x} \cdot \varphi_ {1} (y) + e ^ {- \frac {1 - \sqrt {5}}{2} x} \cdot \varphi_ {2} (y) - \frac {4}{5} e ^ {x + y} (3 \sin (x + y) + 2 \cos (x + y)) \\
\text{where } \varphi_ {1} \text{ and } \varphi_ {2} \text{ are arbitrary functions}
\end{array} \right. { z ( x , y ) = e − 2 1 + 5 x ⋅ φ 1 ( y ) + e − 2 1 − 5 x ⋅ φ 2 ( y ) − 5 4 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y )) where φ 1 and φ 2 are arbitrary functions
2)
( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) z = 4 + 3 x + 6 y (D + D' - 1)(D + 2D' - 3)z = 4 + 3x + 6y ( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) z = 4 + 3 x + 6 y
1 STEP: Let find C.F.
{ ( D + D ′ − 1 ) z ( b D − a D ′ − c ) z → { b = 1 a = − 1 c = 1 → ( C . F . ) 1 = e ( 1 ⋅ x 1 ) ⋅ φ 1 ( 1 ⋅ y + ( − 1 ) ⋅ x ) → \left\{ \begin{array}{l} (D + D' - 1)z \\ (bD - aD' - c)z \end{array} \right. \to \left\{ \begin{array}{l} b = 1 \\ a = -1 \\ c = 1 \end{array} \right. \to (C.F.)_1 = e^{\left(\frac{1 \cdot x}{1}\right)} \cdot \varphi_1(1 \cdot y + (-1) \cdot x) \to { ( D + D ′ − 1 ) z ( b D − a D ′ − c ) z → ⎩ ⎨ ⎧ b = 1 a = − 1 c = 1 → ( C . F . ) 1 = e ( 1 1 ⋅ x ) ⋅ φ 1 ( 1 ⋅ y + ( − 1 ) ⋅ x ) → ( C . F . ) 1 = e x ⋅ φ 1 ( y − x ) , where φ 1 is arbitrary function (C.F.)_1 = e^x \cdot \varphi_1(y - x), \text{ where } \varphi_1 \text{ is arbitrary function} ( C . F . ) 1 = e x ⋅ φ 1 ( y − x ) , where φ 1 is arbitrary function { ( D + 2 D ′ − 3 ) z ( b D − a D ′ − c ) z → { b = 1 a = − 2 c = 3 → ( C . F . ) 2 = e ( 3 ⋅ x 1 ) ⋅ φ 2 ( 1 ⋅ y + ( − 2 ) ⋅ x ) → \left\{ \begin{array}{l} (D + 2D' - 3)z \\ (bD - aD' - c)z \end{array} \right. \to \left\{ \begin{array}{l} b = 1 \\ a = -2 \\ c = 3 \end{array} \right. \to (C.F.)_2 = e^{\left(\frac{3 \cdot x}{1}\right)} \cdot \varphi_2(1 \cdot y + (-2) \cdot x) \to { ( D + 2 D ′ − 3 ) z ( b D − a D ′ − c ) z → ⎩ ⎨ ⎧ b = 1 a = − 2 c = 3 → ( C . F . ) 2 = e ( 1 3 ⋅ x ) ⋅ φ 2 ( 1 ⋅ y + ( − 2 ) ⋅ x ) → ( C . F . ) 2 = e 3 x ⋅ φ 2 ( y − 2 x ) , where φ 2 is arbitrary function (C.F.)_2 = e^{3x} \cdot \varphi_2(y - 2x), \text{ where } \varphi_2 \text{ is arbitrary function} ( C . F . ) 2 = e 3 x ⋅ φ 2 ( y − 2 x ) , where φ 2 is arbitrary function
Then,
C . F . = ( C . F . ) 1 + ( C . F . ) 2 → C . F . = e x ⋅ φ 1 ( y − x ) + e 3 x ⋅ φ 2 ( y − 2 x ) C.F. = (C.F.)_1 + (C.F.)_2 \to C.F. = e^x \cdot \varphi_1(y - x) + e^{3x} \cdot \varphi_2(y - 2x) C . F . = ( C . F . ) 1 + ( C . F . ) 2 → C . F . = e x ⋅ φ 1 ( y − x ) + e 3 x ⋅ φ 2 ( y − 2 x )
2 STEP: Let find P.I.
P . I . = 1 ( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) ( 4 + 3 x + 6 y ) = = 1 ( − 1 ) ⋅ ( 1 − [ D + D ′ ] ) ⋅ ( − 3 ) ⋅ ( 1 − [ D + 2 D ′ 3 ] ) ( 4 + 3 x + 6 y ) = = 1 3 ⋅ ( 1 − [ D + D ′ ] ) − 1 ⋅ ( 1 − [ D + 2 D ′ 3 ] ) − 1 ( 4 + 3 x + 6 y ) = = [ 1 1 − x = 1 + x + x 2 + x 3 + x 4 + ⋯ , ∣ x ∣ < 1 ] = = 1 3 ⋅ ( 1 + [ D + D ′ ] + [ D + D ′ ] 2 + ⋯ ) ⋅ ( 1 + [ D + 2 D ′ 3 ] + [ D + 2 D ′ 3 ] 2 + ⋯ ) ( 4 + 3 x + 6 y ) = 1 3 ⋅ ( 1 + [ D + D ′ ] + [ D + 2 D ′ 3 ] + ⋯ ) ( 4 + 3 x + 6 y ) = = 1 3 ⋅ ( 1 + D + D ′ + D 3 + 2 D ′ 3 ) ( 4 + 3 x + 6 y ) = 1 3 ( 1 + 4 D 3 + 5 D ′ 3 ) ( 4 + 3 x + 6 y ) = \begin{aligned}
P.I. &= \frac{1}{(D + D' - 1)(D + 2D' - 3)} (4 + 3x + 6y) = \\
&= \frac{1}{(-1) \cdot (1 - [D + D']) \cdot (-3) \cdot \left(1 - \left[ \frac{D + 2D'}{3} \right]\right)} (4 + 3x + 6y) = \\
&= \frac{1}{3} \cdot (1 - [D + D'])^{-1} \cdot \left(1 - \left[ \frac{D + 2D'}{3} \right]\right)^{-1} (4 + 3x + 6y) = \\
&= \left[ \frac{1}{1 - x} = 1 + x + x^2 + x^3 + x^4 + \cdots, \, |x| < 1 \right] = \\
&= \frac{1}{3} \cdot (1 + [D + D'] + [D + D']^2 + \cdots) \cdot \left(1 + \left[ \frac{D + 2D'}{3} \right] + \left[ \frac{D + 2D'}{3} \right]^2 + \cdots\right) (4 + 3x + 6y) \\
&= \frac{1}{3} \cdot \left(1 + [D + D'] + \left[ \frac{D + 2D'}{3} \right] + \cdots\right) (4 + 3x + 6y) = \\
&= \frac{1}{3} \cdot \left(1 + D + D' + \frac{D}{3} + \frac{2D'}{3}\right) (4 + 3x + 6y) = \frac{1}{3} \left(1 + \frac{4D}{3} + \frac{5D'}{3}\right) (4 + 3x + 6y) = \\
\end{aligned} P . I . = ( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) 1 ( 4 + 3 x + 6 y ) = = ( − 1 ) ⋅ ( 1 − [ D + D ′ ]) ⋅ ( − 3 ) ⋅ ( 1 − [ 3 D + 2 D ′ ] ) 1 ( 4 + 3 x + 6 y ) = = 3 1 ⋅ ( 1 − [ D + D ′ ] ) − 1 ⋅ ( 1 − [ 3 D + 2 D ′ ] ) − 1 ( 4 + 3 x + 6 y ) = = [ 1 − x 1 = 1 + x + x 2 + x 3 + x 4 + ⋯ , ∣ x ∣ < 1 ] = = 3 1 ⋅ ( 1 + [ D + D ′ ] + [ D + D ′ ] 2 + ⋯ ) ⋅ ( 1 + [ 3 D + 2 D ′ ] + [ 3 D + 2 D ′ ] 2 + ⋯ ) ( 4 + 3 x + 6 y ) = 3 1 ⋅ ( 1 + [ D + D ′ ] + [ 3 D + 2 D ′ ] + ⋯ ) ( 4 + 3 x + 6 y ) = = 3 1 ⋅ ( 1 + D + D ′ + 3 D + 3 2 D ′ ) ( 4 + 3 x + 6 y ) = 3 1 ( 1 + 3 4 D + 3 5 D ′ ) ( 4 + 3 x + 6 y ) = = 1 3 ⋅ ( 4 + 3 x + 6 y + 4 3 ⋅ ∂ ∂ x ( 4 + 3 x + 6 y ) + 5 3 ⋅ ∂ ∂ y ( 4 + 3 x + 6 y ) ) = = 1 3 ⋅ ( 4 + 3 x + 6 y + 4 3 ⋅ 3 + 5 3 ⋅ 6 ) = 1 3 ⋅ ( 4 + 3 x + 6 y + 4 + 10 ) = = 1 3 ⋅ ( 18 + 3 x + 6 y ) = 6 + x + 2 y \begin{array}{l}
= \frac{1}{3} \cdot \left(4 + 3x + 6y + \frac{4}{3} \cdot \frac{\partial}{\partial x} (4 + 3x + 6y) + \frac{5}{3} \cdot \frac{\partial}{\partial y} (4 + 3x + 6y)\right) = \\
= \frac{1}{3} \cdot \left(4 + 3x + 6y + \frac{4}{3} \cdot 3 + \frac{5}{3} \cdot 6\right) = \frac{1}{3} \cdot (4 + 3x + 6y + 4 + 10) = \\
= \frac{1}{3} \cdot (18 + 3x + 6y) = 6 + x + 2y
\end{array} = 3 1 ⋅ ( 4 + 3 x + 6 y + 3 4 ⋅ ∂ x ∂ ( 4 + 3 x + 6 y ) + 3 5 ⋅ ∂ y ∂ ( 4 + 3 x + 6 y ) ) = = 3 1 ⋅ ( 4 + 3 x + 6 y + 3 4 ⋅ 3 + 3 5 ⋅ 6 ) = 3 1 ⋅ ( 4 + 3 x + 6 y + 4 + 10 ) = = 3 1 ⋅ ( 18 + 3 x + 6 y ) = 6 + x + 2 y
Then,
P . I . = 6 + x + 2 y \boxed{P.I. = 6 + x + 2y} P . I . = 6 + x + 2 y
Conclusion,
z ( x , y ) = C . F . + P . I . = e x ⋅ φ 1 ( y − x ) + e 3 x ⋅ φ 2 ( y − 2 x ) + 6 + x + 2 y z(x, y) = C.F. + P.I. = e^x \cdot \varphi_1(y - x) + e^{3x} \cdot \varphi_2(y - 2x) + 6 + x + 2y z ( x , y ) = C . F . + P . I . = e x ⋅ φ 1 ( y − x ) + e 3 x ⋅ φ 2 ( y − 2 x ) + 6 + x + 2 y { z ( x , y ) = e x ⋅ φ 1 ( y − x ) + e 3 x ⋅ φ 2 ( y − 2 x ) + 6 + x + 2 y where φ 1 and φ 2 are arbitrary functions \left\{
\begin{array}{c}
z(x, y) = e^x \cdot \varphi_1(y - x) + e^{3x} \cdot \varphi_2(y - 2x) + 6 + x + 2y \\
\text{where } \varphi_1 \text{ and } \varphi_2 \text{ are arbitrary functions}
\end{array}
\right. { z ( x , y ) = e x ⋅ φ 1 ( y − x ) + e 3 x ⋅ φ 2 ( y − 2 x ) + 6 + x + 2 y where φ 1 and φ 2 are arbitrary functions ANSWER
1)
( D 2 + D − 1 ) z = 4 e x + y cos ( x + y ) → (D^2 + D - 1)z = 4e^{x + y}\cos(x + y) \rightarrow ( D 2 + D − 1 ) z = 4 e x + y cos ( x + y ) → { z ( x , y ) = e − 1 + 5 2 x ⋅ φ 1 ( y ) + e − 1 − 5 2 x ⋅ φ 2 ( y ) − 4 5 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y ) ) where φ 1 and φ 2 are arbitrary functions \left\{
\begin{array}{c}
z(x, y) = e^{-\frac{1 + \sqrt{5}}{2}x} \cdot \varphi_1(y) + e^{-\frac{1 - \sqrt{5}}{2}x} \cdot \varphi_2(y) - \frac{4}{5}e^{x + y}(3\sin(x + y) + 2\cos(x + y)) \\
\text{where } \varphi_1 \text{ and } \varphi_2 \text{ are arbitrary functions}
\end{array}
\right. { z ( x , y ) = e − 2 1 + 5 x ⋅ φ 1 ( y ) + e − 2 1 − 5 x ⋅ φ 2 ( y ) − 5 4 e x + y ( 3 sin ( x + y ) + 2 cos ( x + y )) where φ 1 and φ 2 are arbitrary functions
2)
( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) z = 4 + 3 x + 6 y → (D + D' - 1)(D + 2D' - 3)z = 4 + 3x + 6y \rightarrow ( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) z = 4 + 3 x + 6 y → { z ( x , y ) = e x ⋅ φ 1 ( y − x ) + e 3 x ⋅ φ 2 ( y − 2 x ) + 6 + x + 2 y where φ 1 and φ 2 are arbitrary functions \left\{
\begin{array}{c}
z(x, y) = e^x \cdot \varphi_1(y - x) + e^{3x} \cdot \varphi_2(y - 2x) + 6 + x + 2y \\
\text{where } \varphi_1 \text{ and } \varphi_2 \text{ are arbitrary functions}
\end{array}
\right. { z ( x , y ) = e x ⋅ φ 1 ( y − x ) + e 3 x ⋅ φ 2 ( y − 2 x ) + 6 + x + 2 y where φ 1 and φ 2 are arbitrary functions
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#339914 on Dec 2023