Thread: Decomposition of 2 fourth order differential equations on 4 equations of second order

1. Decomposition of 2 fourth order differential equations on 4 equations of second order

How to make decomposition of 2 fourth order differential equations on 4 equations of second order, just that!

C1x''''+C2x'''+C3x''+C4x'+C5x-C6y''-C7y=0

D1y''''+D2y'''+D3y''+D4y'+D5y-D6x''-D7x=0

Known constants Ci, Di

Thank you in advance

2. Originally Posted by derdack
How to make decomposition of 2 fourth order differential equations on 4 equations of second order, just that!

C1x''''+C2x'''+C3x''+C4x'+C5x-C6y''-C7y=0

D1y''''+D2y'''+D3y''+D4y'+D5y-D6x''-D7x=0

Known constants Ci, Di

Thank you in advance
Use the substitutions

$v=x'' \quad w=y''$

This gives

$C_1v''+C_2v'+C_3v+C_4x'+C_5x-C_6w-C_7y=0$

$D_1w''+D_2w'+D_3w+D_4y'+D_5y-D_6v-D_7x=0$

$x''=v$

$y''=w$

3. Substitution is ok, but I need 4 equations of second order because I must got the system which is the same with some which I already have. To see which influence have the different constants on some parts of equations. I must got system like this one:

q1''+A1*q1'+A2*q1-A3*q2=0
q2''+B1*q2'+B2*q2-B3*q1=0

I know that my system can not reduce on this one, but I can maybe ignore some of high derivative of functions?

What will be the difference between this two equations generally.

q1''''+a*q1''+b*q1-c*q2=0 and then we ignore q1'''', yields q1''+b1*q1-c1*q2=0, then if I already have this equation q1''+b3*q1-c3*q2=0, I know that the difference first is number of solutions, but I must compare the solutions which is associated with the same case of equations???