Calculate A m and B m and use this to calculate X m 1 m odd 1 m 2 1 1 9 1 25 1

Calculate a m and b m and use this to calculate x m 1

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Calculate A m and B m and use this to calculate: X m =1 ,m odd 1 m 2 = 1 + 1 9 + 1 25 + 1 49 · · · A 0 = R 1 - 1 f ( x ) R 1 - 1 1 = R 1 0 1 2 = 1 2 A m = R 1 - 1 f ( x ) cos( πmx ) R 1 - 1 cos 2 ( πmx ) = R 1 0 cos( πmx ) 1 = sin( πmx ) πm 1 0 =0 (We used the fact that f 0 on ( - 1 , 0) ) B 0 = 0
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MATH 54 - FINAL EXAM - SOLUTIONS 11 B m = R 1 - 1 f ( x ) sin( πmx ) R 1 - 1 sin 2 ( πmx ) = R 1 0 cos( πmx ) 1 = - cos( πmx ) πm 1 0 = - 1 πm (cos( πm ) - 1) = - 1 πm (( - 1) m - 1) (We used the fact that f 0 on ( - 1 , 0) ) Now, using Parseval’s identity, we get: X m =0 A 2 m + B 2 m = Z 1 - 1 ( f ( x )) 2 A 2 0 + B 2 0 + X m =1 A 2 m + B 2 m = Z 1 0 1 1 2 2 + 0 2 + X m =1 0 2 + - 1 πm (( - 1) m - 1) =1 X m =1 1 π 2 m 2 (( - 1) m - 1) 2 =1 - 1 4 = 3 4 X m =1 (( - 1) m - 1) 2 m 2 = 3 π 2 4 And finally, to conclude, notice that ( - 1) m - 1 = 0 if m is even and = 2 if m is odd, hence: X m =1 ,m odd 2 2 m 2 = 3 π 2 4 X m =1 ,m odd 1 m 2 = 3 π 2 4(4) = 3 π 2 16
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12 PEYAM RYAN TABRIZIAN 5. (5 points) Use separation of variables to find the general solution to the following PDE: u xx + u yy = u u (0 , y ) = u (1 , y ) =0 (where u = u ( x, y ) and 0 < x < 1 , 0 < y < 1 ) Hint: You can do this!!! Suppose u ( x, y ) = X ( x ) Y ( y ) . Then plug this into the above equation: ( X ( x ) Y ( y )) xx + ( X ( x ) Y ( y )) yy = X ( x ) Y ( y ) X 00 ( x ) Y ( y ) + X ( x ) Y 00 ( y ) = X ( x ) Y ( y ) And divide all the sides by X ( x ) Y ( y ) : X 00 ( x ) Y ( y ) X ( x ) Y ( y ) + X ( x ) Y 00 ( y ) X ( x ) Y ( y ) =1 X 00 ( x ) X ( x ) + Y 00 ( y ) Y ( y ) =1 X 00 ( x ) X ( x ) =1 - Y 00 ( y ) Y ( y ) = λ Hence: X 00 ( x ) = λX ( x ) (and Y 00 ( y ) = (1 - λ ) Y ( y ) ): And as usual, we get that X (0) = 0 and X (1) = 0 , and if we do the 3 - cases business as usual, we find that: λ = - ( πm ) 2 and X ( x ) = sin( πmx ) ( m = 1 , 2 , · · · ) Now go back to Y 00 ( y ) = (1 - λ ) Y ( y ) = (1 + ( πm ) 2 ) Y ( y ) . The auxiliary equation is r 2 = 1+( πm ) 2 , which gives r = ± p 1 + ( πm ) 2 , and hence: Y ( y ) = f A m e 1+( πm ) 2 y + f B m e - 1+( πm ) 2 y And hence:
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MATH 54 - FINAL EXAM - SOLUTIONS 13 X ( x ) Y ( y ) = f A m e 1+( πm ) 2 y + f B m e - 1+( πm ) 2 y sin( πmx ) And finally take linear combinations: u ( x, y ) = X m =1 f A m e 1+( πm ) 2 y + f B m e - 1+( πm ) 2 y sin( πmx )
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