# Chapt_13 - 13 13 Boundary-Value Problems in Rectangular...

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13 13 Boundary-Value Problems in Rectangular Coordinates EXERCISES 13.1 Separable Partial Differential Equations 1. Substituting u ( x, y )= X ( x ) Y ( y ) into the partial diferential equation yields X 0 Y = XY 0 . Separating variables and using the separation constant λ , where λ 6 = 0, we obtain X 0 X = Y 0 Y = λ. When λ 6 =0 X 0 + λX = 0 and Y 0 + λY so that X = c 1 e λx and Y = c 2 e λy . A particular product solution oF the partial diferential equation is u = = c 3 e λ ( x + y ) 6 . When λ = 0 the diferential equations become X 0 = 0 and Y 0 = 0, so in this case X = c 4 , Y = c 5 , and u = = c 6 . 2. Substituting u ( x, y X ( x ) Y ( y ) into the partial diferential equation yields X 0 Y +3 0 = 0. Separating variables and using the separation constant λ we obtain X 0 3 X = Y 0 Y = λ. When λ 6 X 0 3 λX = 0 and Y 0 + λY so that X = c 1 e 3 λx and Y = c 2 e λy . A particular product solution oF the partial diferential equation is u = = c 3 e λ (3 x y ) . When λ = 0 the diferential equations become X 0 = 0 and Y 0 = 0, so in this case X = c 4 , Y = c 5 , and u = = c 6 . 3. Substituting u ( x, y X ( x ) Y ( y ) into the partial diferential equation yields X 0 Y + 0 = . Separating variables and using the separation constant λ we obtain X 0 X = Y Y 0 Y = λ. 680

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13.1 Separable Partial Diferential Equations Then X 0 + λX = 0 and Y 0 (1 + λ ) Y =0 so that X = c 1 e λx and Y = c 2 e (1+ λ ) y . A particular product solution of the partial diFerential equation is u = XY = c 3 e y + λ ( y x ) . 4. Substituting u ( x, y )= X ( x ) Y ( y ) into the partial diFerential equation yields X 0 Y = 0 + . Separating variables and using the separation constant λ we obtain X 0 X = Y + Y 0 Y = λ. Then X 0 + λX = 0 and y 0 +(1+ λ ) Y so that X = c 1 e λx and Y = c 2 e (1+ λ ) y . A particular product solution of the partial diFerential equation is u = = c 3 e y λ ( x + y ) . 5. Substituting u ( x, y X ( x ) Y ( y ) into the partial diFerential equation yields xX 0 Y = yXY 0 . Separating vari- ables and using the separation constant λ we obtain xX 0 X = yY 0 Y = λ. When λ 6 xX 0 + λX = 0 and 0 + λY so that X = c 1 x λ and Y = c 2 y λ . A particular product solution of the partial diFerential equation is u = = c 3 ( xy ) λ . When λ = 0 the diFerential equations become X 0 = 0 and Y 0 = 0, so in this case X = c 4 , Y = c 5 , and u = = c 6 . 6. Substituting u ( x, y X ( x ) Y ( y ) into the partial diFerential equation yields yX 0 Y + xXY 0 = 0. Separating variables and using the separation constant λ we obtain X 0 xX = Y 0 = λ. When λ 6 X 0 + λxX = 0 and Y 0 λyY so that X = c 1 e λx 2 / 2 and Y = c 2 e λy 2 / 2 . A particular product solution of the partial diFerential equation is u = = c 3 e λ ( x 2 y 2 ) / 2 . 681
13.1 Separable Partial Diferential Equations When λ = 0 the diferential equations become X 0 = 0 and Y 0 = 0, so in this case X = c 4 , Y = c 5 , and u = XY = c 6 . 7. Substituting u ( x, y )= X ( x ) Y ( y ) into the partial diferential equation yields X 0 Y + X 0 Y 0 + 0 =0 0 , which is not separable.

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Chapt_13 - 13 13 Boundary-Value Problems in Rectangular...

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