# Hw 3 - Math 445 Solutions to selected problems from HW 3...

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Math 445 - Solutions to selected problems from HW 3 Exercise 1, page 537. Solve for u ( x, y ): u yy + 16 u = 0. Solution We solve the characteristic equation 1 λ 2 + 16 = 0 to obtain λ 1 , 2 = ± 4 i . Thus the general solution is given by a linear combination of cos(4 y ) and sin(4 y ), namely u ( x, y ) = c 1 ( x ) cos(4 y ) + c 2 ( x ) sin(4 y ) . Exercise 5, page 537. Solve for u ( x, y ): u y + u = e xy . Solution By multiplying the above equation with the integrating factor 2 e y we obtain that e y u y + e y u = e ( x +1) y . Now notice that the left hand side of the previous equation is equal to d dy ( e y u ). Thus by integration in y we get e y u = R e ( x +1) y dy + c ( x ) = 1 x +1 e ( x +1) y + c ( x ). Now to obtain u we multiply the previous by e - y and so u ( x, y ) = 1 x +1 e xy + c ( x ) e - y . Exercise 18, page 537. Verify that the given function is a solution of the heat equation with suitable c : u ( x, t ) = e - 2 kt cos(8 x ). Solution

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## This note was uploaded on 02/27/2008 for the course MATH 445 taught by Professor Friedlander during the Fall '07 term at USC.

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Hw 3 - Math 445 Solutions to selected problems from HW 3...

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