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

[ainster31]

Homework Statement

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

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

Homework Equations

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.

 

[Mark44]

Homework Statement

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

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

Homework Equations

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.

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 .