# Thread: System of two 4th order diff equations

1. ## System of two 4th order diff equations

Some ideas for particular solutions? How to make two Wronskian?

a1*x''''[t] + b1*x''[t] + c1*y''''[t] + d1*y''[t] + e1*y[t] == f1[t],
a2*y''''[t] + b2*y''[t] + c2*x''''[t] + d2*x''[t] + e2*x[t] == f2[t],

All constants are known: a1, b1,...

2. Originally Posted by derdack
Some ideas for particular solutions? How to make two Wronskian?

a1*x''''[t] + b1*x''[t] + c1*y''''[t] + d1*y''[t] + e1*y[t] == f1[t],
a2*y''''[t] + b2*y''[t] + c2*x''''[t] + d2*x''[t] + e2*x[t] == f2[t],

All constants are known: a1, b1,...

First you need to use algebra so isolate the highest order derivatives of $x$ and y.

This gives

$\displaystyle x^{(4)}=-\frac{(a_2+b_2)}{c_2}y''-\frac{e_2}{c_2}x-\frac{d_2}{c_2}x''+\frac{f_2}{c_2}$

and

$\displaystyle y^{(4)}=\frac{a_1e_2}{c_1c_2}x-\frac{a_1d_2-b_2c_2}{c_1c_2}x''- \frac{e_1}{c_1}y + \frac{a_1a_2+a_1b_2-d_1c_2}{c_1c_2}y''+\frac{c_2f_1-a_1f_2}{c_1c_2}$

Now use the substitutions

$\displaystyle x'=x_1,x''=x_2,x'''=x_3,\frac{dx_3}{dt}=x^{(4)}$ and
$\displaystyle y'=y_1,y''=y_2,y'''=y_3,\frac{dy_3}{dt}=y^{(4)}$

Now you can decompose this into a system of 8 1st order linear differential equations and use matrix methods to solve

3. I will try. Thank you a lot!!! If I can not find solutions, help will need.

4. In first line it should be (-a2/c2)*y'''' then we will get y4 = ((-a1 b2 + c2 d1) y'')/(a1 a2 - c1 c2) + (c2 e1 y)/(a1 a2 - c1 c2) + ((b1 c2 - a1 d2) x'')/(a1 a2 - c1 c2) - (a1 e2 x)/(a1 a2 -c1 c2) + (a1 f2)/(a1 a2 - c1 c2). How further? We can use substitutions and we must get 8 equations?

5. Originally Posted by derdack
In first line it should be (-a2/c2)*y'''' then we will get y4 = ((-a1 b2 + c2 d1) y'')/(a1 a2 - c1 c2) + (c2 e1 y)/(a1 a2 - c1 c2) + ((b1 c2 - a1 d2) x'')/(a1 a2 - c1 c2) - (a1 e2 x)/(a1 a2 -c1 c2) + (a1 f2)/(a1 a2 - c1 c2). How further? We can use substitutions and we must get 8 equations?
These are the 8 equations

$\displaystyle x'=x_1,x''=x_2,x'''=x_3,\frac{dx_3}{dt}=x^{(4)}$ and
$\displaystyle y'=y_1,y''=y_2,y'''=y_3,\frac{dy_3}{dt}=y^{(4)}$

I will do a bit of the x's

1st equation $\displaystyle \frac{dx}{dt}=x_1$
2nd equation $\displaystyle \frac{d^2x}{dt^2}=x_2 \iff \frac{d}{dt}\left(\frac{dx}{dt} \right)=x_2 \iff \frac{d}{dt}x_1=x_2$
The last step used equation 1
3rd equation $\displaystyle \frac{d^3x}{dt^3}=x_3 \iff \frac{d}{dt}\left(\frac{d^2x}{dt^2} \right)=x_2 \iff \frac{d}{dt}x_2=x_3$
the forth x equations comes from the algebra that I did in the first post

$\displaystyle x^{(4)}=\frac{dx_3}{dt}=-\frac{(a_2+b_2)}{c_2}y_2-\frac{e_2}{c_2}x_1-\frac{d_2}{c_2}x_2+\frac{f_2}{c_2}$

This is the first 4 equations, Now find the other 4 y equations in the exact same way then write the system as

$\mathbf{X}'=A\mathbf{X}+\mathbf{F}(t)$

6. Great!!! Congratulations...

7. In fourth equation when we use the substitution (e2/c2)*x1 where x1=dx/dt. x1 is not correct because it should be (e2/c2)*x in equation? What I should to put instead of x or y to make for equation? I understand that in fourth equation figure x'', y'',x and y. For x''=x2, for y''=y, but what to put for x and y without differentiation?

8. Btw I got x'''' and y'''' after some algebra from first equations in the following form:

x'''' = (a2 f1)/(a1 a2 - c1 c2) - (c1 f2)/(a1 a2 - c1 c2) + ((b2 c1 - a2 e1)/(a1 a2 - c1 c2)) y - ((a2 d1 )/(a1 a2 - c1 c2)) y'' + ((-a2 b1 + c1 e2)/(a1 a2 - c1 c2)) x + ((c1 d2 )/(a1 a2 - c1c2)) x''

y'''' = (c2 f1)/(-a1 a2 + c1 c2) - (a1 f2)/(-a1 a2 +c1 c2) + ((-a1 b2 + c2 e1 )/(a1 a2 - c1 c2)) y -(c2 d1 )/(-a1 a2 + c1 c2)y'' + (((b1 c2 - a1 e2) )/(a1 a2 - c1 c2)) x + ((a1 d2)/(-a1 a2 + c1 c2)) x''

And after that what to put in fourth and 8th diff equation (first order) instead x and y? I understand what to put instead x'' and y''. But what for x and y?

9. Originally Posted by derdack
but what to put for x and y without differentiation?
Yes You are correct that there is a typo in the equaiton. It should read

$\displaystyle x^{(4)}=\frac{dx_3}{dt}=-\frac{(a_2+b_2)}{c_2}y_2-\frac{e_2}{c_2}x-\frac{d_2}{c_2}x_2+\frac{f_2}{c_2}$

Originally Posted by derdack
but what to put for x and y without differentiation?
You don't need to put anything they stay in the equations.

Your variables in the columns will look like

$\mathbf{X}=\begin{pmatrix} x \\ x_1 \\ x_2 \\x_3 \\ y \\ y_1 \\ y_2 \\ y_3\end{pmatrix}$

The matrix $A$ will have 8 rows and 8 columns $8 \times 8$ and only the 4th and 8th rows will have more than one entry

As above this will give

$\mathbf{X}'=A\mathbf{X}+\mathbf{F}(t)$

10. I will try. Ok

11. We didn't get the same equations after algebra. But it is not a problem. I got following:

x'''' = (a2 f1)/(a1 a2 - c1 c2) - (c1 f2)/(a1 a2 - c1 c2) + ((b2 c1 - a2 e1)/(a1 a2 - c1 c2)) y - ((a2 d1 )/(a1 a2 - c1 c2)) y'' + ((-a2 b1 + c1 e2)/(a1 a2 - c1 c2)) x + ((c1 d2 )/(a1 a2 - c1c2)) x''

y'''' = (c2 f1)/(-a1 a2 + c1 c2) - (a1 f2)/(-a1 a2 +c1 c2) + ((-a1 b2 + c2 e1 )/(a1 a2 - c1 c2)) y -(c2 d1 )/(-a1 a2 + c1 c2)y'' + (((b1 c2 - a1 e2) )/(a1 a2 - c1 c2)) x + ((a1 d2)/(-a1 a2 + c1 c2)) x''

I will try to solve with this equations. You help me a lot anyway. If it is not a problem, write me the complete form of matrix equation. X'=AX+F(t). In other problem with variation constants we use wronksian. I am afraid to make a mistake in matrix equation. Thank you anyway!

12. I put the system into Maple and it seems to have given me the same solution that you have. I should have used it in the first place!

The system should look like this

$\displaystyle \frac{d}{dt}\begin{pmatrix} x \\ x_1 \\ x_2 \\x_3 \\ y \\ y_1 \\ y_2 \\ y_3\end{pmatrix}=
\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
\frac{-c_1e_2}{c_1c_2-a_1a_2} & 0 & \frac{a_2b_1-c_1d_2}{c_1c_2-a_1a_2} & 0 &\frac{a_2e_1}{c_1c_2-a_1a_2} & 0 & \frac{a_2d_1-c_1b_2}{c_1c_2-a_1a_2} & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\frac{a_1e_2}{c_1c_2-a_1a_2} & 0 & \frac{a_1d_2-c_2b_1}{c_1c_2-a_1a_2} & 0 &\frac{-c_2e_1}{c_1c_2-a_1a_2} & 0 & \frac{a_1b_2-c_2d_1}{c_1c_2-a_1a_2} & 0 \end{pmatrix}\begin{pmatrix} x \\ x_1 \\ x_2 \\x_3 \\ y \\ y_1 \\ y_2 \\ y_3\end{pmatrix}+\begin{pmatrix} 0 \\ 0 \\ 0 \\ \frac{c_1f_2-a_2f_1}{c_1c_2-a_1a_2}\\ 0 \\ 0 \\ 0 \\ \frac{c_2f_1-a_1f_2}{c_1c_2-a_1a_2}\end{pmatrix}$

Make sure that you double check this

13. Thank you for matrix equation in complete form. That it is understandable. But from initial equations I solve the equations where x'''' and y'''' are unknown, also put in Mathematica software and got x'''' in function of (x'',x) and y'''' in function of (y'',y). Mr Set, you got from second equation x'''' where I can see (a2+b2)/c2 , but a2 multiplied y'''', and I can't see y''''. Anyway thank you a lot for matrix and everything.

14. Originally Posted by derdack
Thank you for matrix equation in complete form. That it is understandable. But from initial equations I solve the equations where x'''' and y'''' are unknown, also put in Mathematica software and got x'''' in function of (x'',x) and y'''' in function of (y'',y). Mr Set, you got from second equation x'''' where I can see (a2+b2)/c2 , but a2 multiplied y'''', and I can't see y''''. Anyway thank you a lot for matrix and everything.
I am not completely sure what you are asking but I think this is it.

On the left hand side we are taking the derivative this gives
for example
$\displaystyle \frac{dx_3}{dt}=$ the bottom row of the matrix remember that

$\displaystyle \frac{dx_3}{dt}=\frac{d^4x}{dt^4}$

So the 4th derivatives are only on the left hand side

15. It is completely clear. But first step from initial equations. It should be to express the x'''' and y'''' from initial equation. I am not sure that it is correct in first your post where you express x'''' and y'''' after some algebra.

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