ANSWER on Question #76084 – Math – Differential Equations
QUESTION
Solve the partial differential equation
( 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
Let us use some 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), \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), \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), \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 + D ′ − 1 ) ( D + 2 D ′ − 3 ) z = 4 + 3 x + 6 y → z ( x , y ) = C . F . + P . I . (D + D' - 1)(D + 2D' - 3)z = 4 + 3x + 6y \rightarrow z(x,y) = C.F. + P.I. ( D + D ′ − 1 ) ( D + 2 D ′ − 3 ) z = 4 + 3 x + 6 y → z ( x , y ) = C . F . + P . I .
1 STEP: Let find C.F.
{ ( D + D ′ − 1 ) z ( b D − a D ′ − c ) z → { b = 1 a = − 1 → ( C . F . ) 1 = e ( 1 ⋅ x 1 ) ⋅ φ 1 ( 1 ⋅ y + ( − 1 ) ⋅ x ) → c = 1 \left\{
\begin{array}{l}
(D + D' - 1)z \\
(bD - aD' - c)z
\end{array}
\right.
\rightarrow
\left\{
\begin{array}{l}
b = 1 \\
a = -1 \rightarrow (C.F.)_1 = e^{\left(\frac{1 \cdot x}{1}\right)} \cdot \varphi_1(1 \cdot y + (-1) \cdot x) \rightarrow \\
c = 1
\end{array}
\right. { ( D + D ′ − 1 ) z ( b D − a D ′ − c ) z → ⎩ ⎨ ⎧ b = 1 a = − 1 → ( C . F . ) 1 = e ( 1 1 ⋅ x ) ⋅ φ 1 ( 1 ⋅ y + ( − 1 ) ⋅ x ) → c = 1 ( 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 . F . ) 2 = e ( 3 ⋅ x 1 ) ⋅ φ 2 ( 1 ⋅ y + ( − 2 ) ⋅ x ) → c = 3 \left\{
\begin{array}{l}
(D + 2D' - 3)z \\
(bD - aD' - c)z
\end{array}
\right.
\rightarrow
\left\{
\begin{array}{l}
b = 1 \\
a = -2 \rightarrow (C.F.)_2 = e^{\left(\frac{3 \cdot x}{1}\right)} \cdot \varphi_2(1 \cdot y + (-2) \cdot x) \rightarrow \\
c = 3
\end{array}
\right. { ( D + 2 D ′ − 3 ) z ( b D − a D ′ − c ) z → ⎩ ⎨ ⎧ b = 1 a = − 2 → ( C . F . ) 2 = e ( 1 3 ⋅ x ) ⋅ φ 2 ( 1 ⋅ y + ( − 2 ) ⋅ x ) → c = 3 ( 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 \rightarrow \boxed{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 ) \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)
\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 ) = 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 ) = = 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(1 + \left[ D + D ^ {\prime} \right] + \left[ \frac {D + 2 D ^ {\prime}}{3} \right] + \dots\right) (4 + 3 x + 6 y) = \\
= \frac {1}{3} \cdot \left(1 + D + D ^ {\prime} + \frac {D}{3} + \frac {2 D ^ {\prime}}{3}\right) (4 + 3 x + 6 y) = \frac {1}{3} \left(1 + \frac {4 D}{3} + \frac {5 D ^ {\prime}}{3}\right) (4 + 3 x + 6 y) = \\
= \frac {1}{3} \cdot \left(4 + 3 x + 6 y + \frac {4}{3} \cdot \frac {\partial}{\partial x} (4 + 3 x + 6 y) + \frac {5}{3} \cdot \frac {\partial}{\partial y} (4 + 3 x + 6 y)\right) = \\
= \frac {1}{3} \cdot \left(4 + 3 x + 6 y + \frac {4}{3} \cdot 3 + \frac {5}{3} \cdot 6\right) = \frac {1}{3} \cdot (4 + 3 x + 6 y + 4 + 1 0) = \\
= \frac {1}{3} \cdot (1 8 + 3 x + 6 y) = 6 + x + 2 y \\
\end{array} = 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 ) = = 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 + 2 y} 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 ^ {3 x} \cdot \varphi_ {2} (y - 2 x) + 6 + x + 2 y 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 ^ {3 x} \cdot \varphi_ {2} (y - 2 x) + 6 + x + 2 y \\ \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:
{ 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 ^ {3 x} \cdot \varphi_ {2} (y - 2 x) + 6 + x + 2 y \\ \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