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# 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 differential equation yields X Y = XY . Separating variables and using the separation constant λ , where λ = 0, we obtain X X = Y Y = λ. When λ = 0 X + λX = 0 and Y + λY = 0 so that X = c 1 e λx and Y = c 2 e λy . A particular product solution of the partial differential equation is u = XY = c 3 e λ ( x + y ) , λ = 0 . When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c 4 , Y = c 5 , and u = XY = c 6 . 2. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields X Y + 3 XY = 0. Separating variables and using the separation constant λ we obtain X 3 X = Y Y = λ. When λ = 0 X 3 λX = 0 and Y + λY = 0 so that X = c 1 e 3 λx and Y = c 2 e λy . A particular product solution of the partial differential equation is u = XY = c 3 e λ (3 x y ) . When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c 4 , Y = c 5 , and u = XY = c 6 . 3. Substituting u ( x, y ) = X ( x ) Y ( y ) into the partial differential equation yields X Y + XY = XY . Separating variables and using the separation constant λ we obtain X X = Y Y Y = λ. 680

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

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