# 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

$$C_1x''''+C_2x'''+C_3x''+C_4x'+C_5x-C_6y''-C_7y=0$$

$$D_1y''''+D_2y'''+D_3y''+D_4y'+D_5y-D_6x''-D_7x=0$$

Known constants $$C_i, D_i$$

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Hint:

Define $w=x''$ and $v=y''$.

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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? –  Derdack May 7 '11 at 18:36
@Derdack: maybe I don't understand you completely: you get 4 equations. 2 are given as a hint. The other two are in your post. –  Fabian May 7 '11 at 18:38
Write your system in the matrix form $\bf{z}^{\prime}=A\mathbf{z},$ where $\bf{z}=\left( x,x^{\prime},x^{\prime\prime},x^{\prime\prime\prime },y,y^{\prime},y^{\prime\prime},y^{\prime\prime\prime}\right) ^{T}$ (you can divide the first equation with $C_{1}$ and the second one with $D_{1}%$, assuming these are non-zero). Now try to find a transformation $\bf{z=}T\mathbf{q}$ such that $\bf{q}^{\prime}=T^{-1}AT\mathbf{q}$ has the required form (here $\mathbf{q}=\left( q_{1},q_{1}^{\prime},q_{2},q_{2}^{\prime},q_{3},q_{3}^{\prime},q_{4},q_{4}^{\prime}\right) ^{T}$ with your notation). Is this possible?