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### hw_4_14_09

Course: MA MA221, Spring 2009
School: Stevens
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Word Count: 215

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221 MA Homework Solutions Due date: April 14, 2009 Section 10.2, #27 &amp; 29 (Underlined Problems are to be handed in) 27.) 2 u 1 u 1 2 u 0 r r r 2 r 2 2 with ur, RrT. Subsituting into the PDE we get R T 1 R T - 1 RT r r2 or r 2 R rR - T R T since the left hand side is purely a function of r, and the right hand side is purely a function of . Thus we have the two ODEs r 2 R rR - R 0...

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221 MA Homework Solutions Due date: April 14, 2009 Section 10.2, #27 & 29 (Underlined Problems are to be handed in) 27.) 2 u 1 u 1 2 u 0 r r r 2 r 2 2 with ur, RrT. Subsituting into the PDE we get R T 1 R T - 1 RT r r2 or r 2 R rR - T R T since the left hand side is purely a function of r, and the right hand side is purely a function of . Thus we have the two ODEs r 2 R rR - R 0 T T 0 29.) u u xx u yy t with ux, y, t XxYyTt. Substituting into the PDE we get XYT X YT T XY or 1 T X Y XY XY T Since the left hand side depends only on t, whereas the right hand side depends only on x and y, then 1 T X Y XY K XY T where K is a constant. Therefore the equation for Tt is T - KT 0 Now we also have from the above euqation that X Y XY KXY or 1 X - Y K X Y Since the left hand side depends only on x, whereas the right hand side depends only on y, we have X - Y K J X Y where J is a constant. Therefore, we get the equations X - KX 0 Y J - KY 0 2
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