ANSWER on Question #79497 – Math – Differential Equations
QUESTION
Solve the partial differential equation
( D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ ) z = sin ( x + 2 y ) (D^2 + 2D D' + D'^2 - 2D - 2D')z = \sin(x + 2y) ( D 2 + 2 D D ′ + D ′2 − 2 D − 2 D ′ ) z = sin ( x + 2 y ) SOLUTION
Let us use some well-known 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 ) .
In our case,
( D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ ) z = sin ( x + 2 y ) → z ( x , y ) = C . F . + P . I . (D^{2} + 2DD' + D'^{2} - 2D - 2D')z = \sin(x + 2y) \rightarrow z(x,y) = C.F. + P.I. ( D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ ) z = sin ( x + 2 y ) → z ( x , y ) = C . F . + P . I .
0 STEP: We factor the expression
D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ = ( D 2 + 2 D D ′ + D ′ 2 ) + ( − 2 D − 2 D ′ ) = = ( D + D ′ ) 2 − 2 ⋅ ( D + D ′ ) = ( D + D ′ ) ( D + D ′ − 2 ) \begin{aligned}
D^{2} + 2DD' + D'^{2} - 2D - 2D' &= (D^{2} + 2DD' + D'^{2}) + (-2D - 2D') = \\
&= (D + D')^{2} - 2 \cdot (D + D') = (D + D')(D + D' - 2)
\end{aligned} D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ = ( D 2 + 2 D D ′ + D ′ 2 ) + ( − 2 D − 2 D ′ ) = = ( D + D ′ ) 2 − 2 ⋅ ( D + D ′ ) = ( D + D ′ ) ( D + D ′ − 2 )
Conclusion,
( D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ ) z = ( D + D ′ ) ( D + D ′ − 2 ) z \boxed{(D^{2} + 2DD' + D'^{2} - 2D - 2D')z = (D + D')(D + D' - 2)z} ( D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ ) z = ( D + D ′ ) ( D + D ′ − 2 ) z
1 STEP: Let find C.F.
C . F . = ( C . F . ) 1 + ( C . F . ) 2 C.F. = (C.F.)_1 + (C.F.)_2 C . F . = ( C . F . ) 1 + ( C . F . ) 2 { ( D + D ′ ) z ( b D − a D ′ − c ) z → { b = 1 a = − 1 → ( C . F . ) 1 = e ( 0 ⋅ x 1 ) ⋅ φ 1 ( 1 ⋅ y + ( − 1 ) ⋅ x ) → c = 0 \begin{cases}
(D + D')z \\
(bD - aD' - c)z
\end{cases}
\rightarrow
\begin{cases}
b = 1 \\
a = -1 \rightarrow (C.F.)_1 = e^{\left(\frac{0 \cdot x}{1}\right)} \cdot \varphi_1(1 \cdot y + (-1) \cdot x) \rightarrow \\
c = 0
\end{cases} { ( D + D ′ ) z ( b D − a D ′ − c ) z → ⎩ ⎨ ⎧ b = 1 a = − 1 → ( C . F . ) 1 = e ( 1 0 ⋅ x ) ⋅ φ 1 ( 1 ⋅ y + ( − 1 ) ⋅ x ) → c = 0 ( C . F . ) 1 = φ 1 ( y − x ) , where φ 1 is an arbitrary function \boxed{(C.F.)_1 = \varphi_1(y - x), \text{ where } \varphi_1 \text{ is an arbitrary function}} ( C . F . ) 1 = φ 1 ( y − x ) , where φ 1 is an arbitrary function { ( D + D ′ − 2 ) z ( b D − a D ′ − c ) z → { b = 1 a = − 1 → ( C . F . ) 2 = e ( 2 ⋅ x 1 ) ⋅ φ 2 ( 1 ⋅ y + ( − 1 ) ⋅ x ) → c = 2 \begin{cases}
(D + D' - 2)z \\
(bD - aD' - c)z
\end{cases}
\rightarrow
\begin{cases}
b = 1 \\
a = -1 \rightarrow (C.F.)_2 = e^{\left(\frac{2 \cdot x}{1}\right)} \cdot \varphi_2(1 \cdot y + (-1) \cdot x) \rightarrow \\
c = 2
\end{cases} { ( D + D ′ − 2 ) z ( b D − a D ′ − c ) z → ⎩ ⎨ ⎧ b = 1 a = − 1 → ( C . F . ) 2 = e ( 1 2 ⋅ x ) ⋅ φ 2 ( 1 ⋅ y + ( − 1 ) ⋅ x ) → c = 2 ( C . F . ) 2 = e 2 x φ 2 ( y − x ) , where φ 2 is an arbitrary function \boxed{(C.F.)_2 = e^{2x} \varphi_2(y - x), \text{ where } \varphi_2 \text{ is an arbitrary function}} ( C . F . ) 2 = e 2 x φ 2 ( y − x ) , where φ 2 is an arbitrary function
Then,
C . F . = ( C . F . ) 1 + ( C . F . ) 2 → C . F . = φ 1 ( y − x ) + e 2 x φ 2 ( y − x ) C.F. = (C.F.)_1 + (C.F.)_2 \rightarrow \boxed{C.F. = \varphi_1(y - x) + e^{2x} \varphi_2(y - x)} C . F . = ( C . F . ) 1 + ( C . F . ) 2 → C . F . = φ 1 ( y − x ) + e 2 x φ 2 ( y − x )
2 STEP: Let find P.I.
P . I . = 1 D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ sin ( x + 2 y ) = 1 D 2 + 2 D D ′ + D ′ 2 − 2 D − 2 D ′ sin ( 1 ⋅ x + 2 ⋅ y ) = = 1 ( 1 ⋅ i ) 2 + 2 ⋅ ( 1 ⋅ i ) ⋅ ( 2 ⋅ i ) + ( 2 ⋅ i ) 2 − 2 D − 2 D ′ sin ( 1 ⋅ x + 2 ⋅ y ) = = 1 i 2 + 2 ⋅ 2 i 2 + 4 i 2 − 2 D − 2 D ′ sin ( x + 2 y ) = 1 9 i 2 − 2 D − 2 D ′ sin ( x + 2 y ) = = 1 − 9 − 2 D − 2 D ′ sin ( x + 2 y ) = 1 − 1 ⋅ ( 2 D + 2 D ′ + 9 ) sin ( x + 2 y ) = − 1 ( 2 D + 2 D ′ ) + 9 sin ( x + 2 y ) = = − 1 ⋅ ( ( 2 D + 2 D ′ ) − 9 ) ( ( 2 D + 2 D ′ ) + 9 ) ⋅ ( ( 2 D + 2 D ′ ) − 9 ) sin ( x + 2 y ) = 9 − ( 2 D + 2 D ′ ) ( 2 D + 2 D ′ ) 2 − 81 sin ( x + 2 y ) = = 9 − ( 2 D + 2 D ′ ) 4 D 2 + 8 D D ′ + 4 D ′ 2 − 81 sin ( x + 2 y ) = = 9 − ( 2 D + 2 D ′ ) 4 ⋅ ( 1 ⋅ i ) 2 + 8 ⋅ ( 1 ⋅ i ) ⋅ ( 2 ⋅ i ) + 4 ⋅ ( 2 ⋅ i ) 2 − 81 sin ( 1 ⋅ x + 2 ⋅ y ) = = 9 − ( 2 D + 2 D ′ ) 4 i 2 + 16 i 2 + 16 i 2 − 81 sin ( x + 2 y ) = 9 − ( 2 D + 2 D ′ ) 36 i 2 − 81 sin ( x + 2 y ) = 9 − ( 2 D + 2 D ′ ) − 36 − 81 sin ( x + 2 y ) = = 9 − ( 2 D + 2 D ′ ) − 117 sin ( x + 2 y ) = 1 117 ⋅ ( ( 2 D + 2 D ′ ) − 9 ) sin ( x + 2 y ) = = 1 117 ( 2 ⋅ ∂ ∂ x + 2 ⋅ ∂ ∂ y − 9 ) sin ( x + 2 y ) = = 1 117 ( 2 ⋅ ∂ ∂ x ( sin ( x + 2 y ) ) + 2 ⋅ ∂ ∂ y ( sin ( x + 2 y ) ) − 9 ⋅ sin ( x + 2 y ) ) = = 1 117 ( 2 ⋅ cos ( x + 2 y ) + 2 ⋅ 2 ⋅ cos ( x + 2 y ) − 9 ⋅ sin ( x + 2 y ) ) = 6 cos ( x + 2 y ) − 9 sin ( x + 2 y ) 117 = = 3 ⋅ ( 2 cos ( x + 2 y ) − 3 sin ( x + 2 y ) ) 3 ⋅ 39 = 2 cos ( x + 2 y ) − 3 sin ( x + 2 y ) 39 ≡ ≡ 2 39 ⋅ cos ( x + 2 y ) − 3 39 ⋅ sin ( x + 2 y ) \begin{aligned}
P.I. &= \frac{1}{D^2 + 2DD' + D'^2 - 2D - 2D'} \sin(x + 2y) = \frac{1}{D^2 + 2DD' + D'^2 - 2D - 2D'} \sin(1 \cdot x + 2 \cdot y) = \\
&= \frac{1}{(1 \cdot i)^2 + 2 \cdot (1 \cdot i) \cdot (2 \cdot i) + (2 \cdot i)^2 - 2D - 2D'} \sin(1 \cdot x + 2 \cdot y) = \\
&= \frac{1}{i^2 + 2 \cdot 2i^2 + 4i^2 - 2D - 2D'} \sin(x + 2y) = \frac{1}{9i^2 - 2D - 2D'} \sin(x + 2y) = \\
&= \frac{1}{-9 - 2D - 2D'} \sin(x + 2y) = \frac{1}{-1 \cdot (2D + 2D' + 9)} \sin(x + 2y) = \frac{-1}{(2D + 2D') + 9} \sin(x + 2y) = \\
&= \frac{-1 \cdot ((2D + 2D') - 9)}{((2D + 2D') + 9) \cdot ((2D + 2D') - 9)} \sin(x + 2y) = \frac{9 - (2D + 2D')}{(2D + 2D')^2 - 81} \sin(x + 2y) = \\
&= \frac{9 - (2D + 2D')}{4D^2 + 8DD' + 4D'^2 - 81} \sin(x + 2y) = \\
&= \frac{9 - (2D + 2D')}{4 \cdot (1 \cdot i)^2 + 8 \cdot (1 \cdot i) \cdot (2 \cdot i) + 4 \cdot (2 \cdot i)^2 - 81} \sin(1 \cdot x + 2 \cdot y) = \\
&= \frac{9 - (2D + 2D')}{4i^2 + 16i^2 + 16i^2 - 81} \sin(x + 2y) = \frac{9 - (2D + 2D')}{36i^2 - 81} \sin(x + 2y) = \frac{9 - (2D + 2D')}{-36 - 81} \sin(x + 2y) = \\
&= \frac{9 - (2D + 2D')}{-117} \sin(x + 2y) = \frac{1}{117} \cdot ((2D + 2D') - 9) \sin(x + 2y) = \\
&= \frac{1}{117} \left(2 \cdot \frac{\partial}{\partial x} + 2 \cdot \frac{\partial}{\partial y} - 9\right) \sin(x + 2y) = \\
&= \frac{1}{117} \left(2 \cdot \frac{\partial}{\partial x} (\sin(x + 2y)) + 2 \cdot \frac{\partial}{\partial y} (\sin(x + 2y)) - 9 \cdot \sin(x + 2y)\right) = \\
&= \frac{1}{117} (2 \cdot \cos(x + 2y) + 2 \cdot 2 \cdot \cos(x + 2y) - 9 \cdot \sin(x + 2y)) = \frac{6 \cos(x + 2y) - 9 \sin(x + 2y)}{117} = \\
&= \frac{3 \cdot (2 \cos(x + 2y) - 3 \sin(x + 2y))}{3 \cdot 39} = \frac{2 \cos(x + 2y) - 3 \sin(x + 2y)}{39} \equiv \\
&\equiv \frac{2}{39} \cdot \cos(x + 2y) - \frac{3}{39} \cdot \sin(x + 2y)
\end{aligned} P . I . = D 2 + 2 D D ′ + D ′2 − 2 D − 2 D ′ 1 sin ( x + 2 y ) = D 2 + 2 D D ′ + D ′2 − 2 D − 2 D ′ 1 sin ( 1 ⋅ x + 2 ⋅ y ) = = ( 1 ⋅ i ) 2 + 2 ⋅ ( 1 ⋅ i ) ⋅ ( 2 ⋅ i ) + ( 2 ⋅ i ) 2 − 2 D − 2 D ′ 1 sin ( 1 ⋅ x + 2 ⋅ y ) = = i 2 + 2 ⋅ 2 i 2 + 4 i 2 − 2 D − 2 D ′ 1 sin ( x + 2 y ) = 9 i 2 − 2 D − 2 D ′ 1 sin ( x + 2 y ) = = − 9 − 2 D − 2 D ′ 1 sin ( x + 2 y ) = − 1 ⋅ ( 2 D + 2 D ′ + 9 ) 1 sin ( x + 2 y ) = ( 2 D + 2 D ′ ) + 9 − 1 sin ( x + 2 y ) = = (( 2 D + 2 D ′ ) + 9 ) ⋅ (( 2 D + 2 D ′ ) − 9 ) − 1 ⋅ (( 2 D + 2 D ′ ) − 9 ) sin ( x + 2 y ) = ( 2 D + 2 D ′ ) 2 − 81 9 − ( 2 D + 2 D ′ ) sin ( x + 2 y ) = = 4 D 2 + 8 D D ′ + 4 D ′2 − 81 9 − ( 2 D + 2 D ′ ) sin ( x + 2 y ) = = 4 ⋅ ( 1 ⋅ i ) 2 + 8 ⋅ ( 1 ⋅ i ) ⋅ ( 2 ⋅ i ) + 4 ⋅ ( 2 ⋅ i ) 2 − 81 9 − ( 2 D + 2 D ′ ) sin ( 1 ⋅ x + 2 ⋅ y ) = = 4 i 2 + 16 i 2 + 16 i 2 − 81 9 − ( 2 D + 2 D ′ ) sin ( x + 2 y ) = 36 i 2 − 81 9 − ( 2 D + 2 D ′ ) sin ( x + 2 y ) = − 36 − 81 9 − ( 2 D + 2 D ′ ) sin ( x + 2 y ) = = − 117 9 − ( 2 D + 2 D ′ ) sin ( x + 2 y ) = 117 1 ⋅ (( 2 D + 2 D ′ ) − 9 ) sin ( x + 2 y ) = = 117 1 ( 2 ⋅ ∂ x ∂ + 2 ⋅ ∂ y ∂ − 9 ) sin ( x + 2 y ) = = 117 1 ( 2 ⋅ ∂ x ∂ ( sin ( x + 2 y )) + 2 ⋅ ∂ y ∂ ( sin ( x + 2 y )) − 9 ⋅ sin ( x + 2 y ) ) = = 117 1 ( 2 ⋅ cos ( x + 2 y ) + 2 ⋅ 2 ⋅ cos ( x + 2 y ) − 9 ⋅ sin ( x + 2 y )) = 117 6 cos ( x + 2 y ) − 9 sin ( x + 2 y ) = = 3 ⋅ 39 3 ⋅ ( 2 cos ( x + 2 y ) − 3 sin ( x + 2 y )) = 39 2 cos ( x + 2 y ) − 3 sin ( x + 2 y ) ≡ ≡ 39 2 ⋅ cos ( x + 2 y ) − 39 3 ⋅ sin ( x + 2 y )
Conclusion,
\boxed{P.I.} = \frac{2}{39} \cdot \cos(x + 2y) - \frac{3}{39} \cdot \sin(x + 2y)}
General Conclusion,
z ( x , y ) = C . F . + P . I . = φ 1 ( y − x ) + e 2 x φ 2 ( y − x ) + 2 39 ⋅ cos ( x + 2 y ) − 3 39 ⋅ sin ( x + 2 y ) → z(x, y) = C.F. + P.I. = \varphi_1(y - x) + e^{2x} \varphi_2(y - x) + \frac{2}{39} \cdot \cos(x + 2y) - \frac{3}{39} \cdot \sin(x + 2y) \rightarrow z ( x , y ) = C . F . + P . I . = φ 1 ( y − x ) + e 2 x φ 2 ( y − x ) + 39 2 ⋅ cos ( x + 2 y ) − 39 3 ⋅ sin ( x + 2 y ) → \boxed{z(x, y) = \varphi_1(y - x) + e^{2x} \varphi_2(y - x) + \frac{2}{39} \cdot \cos(x + 2y) - \frac{3}{39} \cdot \sin(x + 2y)}}
ANSWER:
{ z ( x , y ) = φ 1 ( y − x ) + e 2 x φ 2 ( y − x ) + 2 39 ⋅ cos ( x + 2 y ) − 3 39 ⋅ sin ( x + 2 y ) where φ 1 and φ 2 are arbitrary functions \left\{ \begin{array}{c} z(x, y) = \varphi_1(y - x) + e^{2x} \varphi_2(y - x) + \frac{2}{39} \cdot \cos(x + 2y) - \frac{3}{39} \cdot \sin(x + 2y) \\ \text{where } \varphi_1 \text{ and } \varphi_2 \text{ are arbitrary functions} \end{array} \right. { z ( x , y ) = φ 1 ( y − x ) + e 2 x φ 2 ( y − x ) + 39 2 ⋅ cos ( x + 2 y ) − 39 3 ⋅ sin ( x + 2 y ) where φ 1 and φ 2 are arbitrary functions
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#339914 on Dec 2023