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What am I doing wrong in attempting to solve this system of differenti

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data

    What am I doing wrong in attempting to solve this system of differential equations?

    Problem:$$\quad x''+y''=t^{ 2 }\quad \quad x''-y''=4t\quad \quad x(0)=8\quad x'(0)=y(0)=y'(0)=0\\ \\$$

    2. Relevant equations



    3. The attempt at a solution

    Attempt:$$\\ s^{ 2 }L\{ x\} -sx(0)-x'(0)+{ s }^{ 2 }L\{ y\} -sy(0)-y'(0)=2/{ s }^{ 3 }\\ { s }^{ 5 }L\{ x\} -8{ s }^{ 4 }+{ s }^{ 5 }L\{ y\} =2\quad \quad (1)\\ \\ { s }^{ 4 }L\{ x\} -8{ s }^{ 3 }-{ s }^{ 4 }L\{ y\} =4\quad \quad (2)\\ \\ (1)-(2)\cdot s:\quad 16s^{ 4 }+2s^{ 5 }L\{ y\} =2-4s\\ L\{ y\} =\frac { -8 }{ s } -\frac { 2 }{ { s }^{ 4 } } +\frac { 1 }{ { s }^{ 5 } } \\ y=-8-\frac { { t }^{ 3 } }{ 3 } +\frac { { t }^{ 4 } }{ 24 } \\ Sub\quad L\{ y\} \quad into\quad (2)\\ L\{ x\} =\frac { 2+1/2 }{ { s }^{ 4 } } \\ x=\frac { { t }^{ 3 } }{ 3 } +\frac { { t }^{ 4 } }{ 24 }$$

    I know my attempt is wrong because x(0)=8 but if I compute x(0), I get 0. My prof's solution is also wrong for the same reason.
     
  2. jcsd
  3. Nov 13, 2013 #2

    Mark44

    Staff: Mentor

    Are you required to use Laplace transforms on this? It turns out to be much simpler by other means.

    Here's what I'm talking about.
    x'' + y'' = t2
    x'' - y'' = 4t

    Add the first equation to the second to get 2x'' = t2 + 4t, or x'' = (1/2)t2 + 2t. This is easy to solve by integrating twice.

    To get an equation involving only y derivatives, subtract the second equation from the first, to get 2y'' = t2 - 4t, or y'' = (1/2)t2 - 2t .
     
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