Linear system of differential equations

[etf]

Hi!
My task is to solve this system:
$$\frac{\mathrm{d} x}{\mathrm{d} t}=-x+y-2z$$
$$\frac{\mathrm{d} y}{\mathrm{d} t}=4x+y$$
$$\frac{\mathrm{d} z}{\mathrm{d} t}=2x+y-z$$
My first equation (1) is $$\frac{\mathrm{d} y}{\mathrm{d} t}=4x+y$$.
Derivative of (1) is $$\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t})=4\frac{\mathrm{d} x}{\mathrm{d} t}+\frac{\mathrm{d} y}{\mathrm{d} t}=...=5y-8z$$ and this is my second equation.
Derivative of my second equation is $$\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t}))=5\frac{\mathrm{d} y}{\mathrm{d} t}-8\frac{\mathrm{d} z}{\mathrm{d} t}=...=4x-3y+8z$$ and this my third equation. So, my three equations are:
$$\frac{\mathrm{d} y}{\mathrm{d} t}=4x+y . . . (1)$$
$$\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t})=5y-8z . . . (2)$$
$$\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t}))=4x-3y+8z . . . (3)$$
From equation (1) we get $$4x=\frac{\mathrm{d} y}{\mathrm{d} t}-y$$ and we substitute it in equation (3):
$$\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t}))=4x-3y+8z=\frac{\mathrm{d} y}{\mathrm{d} t}-y-3y+8z$$
$$\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t}))=\frac{\mathrm{d} y}{\mathrm{d} t}-4y+8z . . . (4)$$
From equation (2) we get $$8z=5y-\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t})$$ and we substitute it in equation (4) so equation (4) becomes:
$$\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t}))+\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t})-\frac{\mathrm{d} y}{\mathrm{d} t}-y=0$$. This is Linear homogeneous differential equation and it has general solution $$y(t)=C1e^{t}+C2te^{-t}+C3t^2e^{-t}.$$
Now we can calculate $$x$$ from equation (1), $$x=\frac{1}{4}(\frac{\mathrm{d} y}{\mathrm{d} t}-y)$$.
We get $$x(t)=\frac{1}{4}C2(e^{-t}-2te^{-t})+\frac{1}{4}C3(2te^{-t}-2t^2e^{-t}).$$
From first equation of our system we get:
$$z(t)=-\frac{1}{2}\frac{\mathrm{d} x}{\mathrm{d} t}-\frac{1}{2}x+\frac{1}{2}y=...=\frac{1}{2}C1e^{t}+C2(\frac{1}{4}e^{-t}+\frac{1}{2}te^{-t})-C3(\frac{1}{4}e^{-t}-\frac{1}{2}te^{-t}-\frac{1}{2}t^2e^{-t})$$
I checked it and I couldn't figure out what's wrong with my solution. I used Matlab to check my solution. I used simplify(diff(x,t)+x-y+2*z) and simplify(diff(y,t)-4*x-y)to check my solution. I got 0 result for both commands and I thought everything is fine but I tried simplify(diff(z,t)-2*x-y+z) and I got (C3*exp(-t))/2

 

[Simon Bridge]

I have been unable to follow your algebra.
Try doing it in terms of eigenvalues...
http://tutorial.math.lamar.edu/Classes/DE/SystemsDE.aspx

 

[etf]

This is how my teacher told me to solve systems of D.E. I will take a look at your method.

 

[verty]

Or look here.

 

[ehild]

$$\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t}))+\frac{\mathrm{d} }{\mathrm{d} t}(\frac{\mathrm{d} y}{\mathrm{d} t})-\frac{\mathrm{d} y}{\mathrm{d} t}-y=0$$. This is Linear homogeneous differential equation and it has general solution $$y(t)=C1e^{t}+C2te^{-t}+C3t^2e^{-t}.$$

No, the general solution is not that.

Try solutions in the form e^λt. Substitute into the differential equation, you get an algebraic equation for λ. If two roots are the same, (λ1) and the third one different, (λ2) then one solution belonging to λ₁ is e^λ₁t and the other is te^λ₁t. But the solution that belongs to λ₂ is e^λ₂t.

So the general solution for y is y(t)=C₁e^-t+C₂te^-t+C₃e^t.

ehild

 

[etf]

@ehild,
you saved me! :) thanks a lot!

 

[ehild]

You are welcome.

It would be really much less work if you start with the trial solution from the beginning. Assume all variables in the form Ce^λt, but with different constant coefficients:

$x(t)=X_0 e^{λt}$; $y(t)=Y_0 e^{λt}$; $z(t)=Z_0 e^{λt}$.

Substitute into the original system of equations. You get a linear homogeneous system for X₀, Y₀, Z₀. I think, you are familiar with such equations and you know that it has non-zero solutions if the determinant of the coefficients is zero. Solving the determinant equation, you get the values for λ. Substituting the roots back, you can determine X₀, Y₀, Z₀ (one of them free) for each of them.

ehild

 

[etf]

I have one more question. Here is my system:

$$\\\frac{\mathrm{d} x}{\mathrm{d} t}=5x+6y-6z$$
$$\\\frac{\mathrm{d} y}{\mathrm{d} t}=-2x-3y+4z$$
$$\\\frac{\mathrm{d} z}{\mathrm{d} t}=x+y$$

Here is my solution:
$$\\z'=x+y...(1)$$
$$\\z''=x'+y'=5x+6y-6z-2x-3y+4z=3x+3y-2z$$
$$\\z''=3x+3y-2z...(2)$$
$$\\z'''=3x'+3y'-2z'=...$$
$$\\z'''=7x+7y-6z...(3)$$

From equation (1) we get:
$$\\x=z'-y$$
We substitute it in equation (2) and we get:
$$\\z''=3x+3y-2z=3z'-3y+3y-2z$$
$$\\z''-3z'+2z=0\\$$
This is linear homogeneous D.E. with solution $$\\z(t)=C1e^{2t}+C2e^{t}$$.
We multiply last equation from our system of equations with (-6) and we add it to our first equation (also from our system). We get:
$$\\x'-6z'=-6x+5x-6y+6y-6z$$
$$\\x'+x=6z'-6z=6(2C1e^{2t}+C2e^{t})-6C1e^{2t}-6C2e^{t}$$
$$\\x'+x=6C1e^{2t}$$
We will solve this linear D.E. using method of variation of constant.
$$\\x'+x=0$$
$$\\\frac{\mathrm{d} x}{\mathrm{d} t}=-x/*\frac{dt}{x}$$
$$\\\frac{dx}{x}=-dt$$
$$\\lnx-lnC=-t$$
$$\\ln\frac{x}{C}=-t$$
$$\\x=Ce^{-t}$$
$$\\C=\varphi (t) \rightarrow x=\varphi (t)e^{-t}$$
$$\\x'=\varphi (t)'e^{-t}-\varphi (t)e^{-t}$$
$$\\\varphi (t)'e^{-t}-\varphi (t)e^{-t}+\varphi (t)e^{-t}=6C1e^{2t}$$
$$\\\varphi (t)'e^{-t}=6C1e^{2t}$$
$$\\\varphi (t)'=6C1e^{3t}$$
$$\\\varphi (t)=2C1e^{3t}+C3$$
Now we can calculate x(t):
$$\\x(t)=Ce^{-t}=\varphi (t)e^{-t}=(2C1e^{3t}+C3)e^{-t}$$
It is easy now to calculate y(t) as y(t)=z'-x. I checked my solution and it's ok but my question is, why it is necessary to add constant C3 when solving for $$\varphi(t)$$ (IT MUST BE C3, NO C1 or C2) ? If I use C1 or C2 instead of C3 I GOT WRONG SOLUTION

 

[Simon Bridge]

why it is necessary to add constant C3 when solving for
φ(t)
(IT MUST BE C3, NO C1 or C2) ? If I use C1 or C2 instead of C3 I GOT WRONG SOLUTION

... because someone didn't program the computer properly, probably.
It does not matter what you call the arbitrary constant - but a computer marking program does not know that unless someone tells it.

Sometimes a human marker will use the variable name as an indicator to how you are thinking - but that would be poor teaching.

 

[ehild]

You have used C1 and C2 for z(t). A new integration constant is needed for φ(t).

The method you are forced to follow is terribly complicated with respect to the standard method, which gives the three roots (2, 1,-1) in two lines.

ehild

 

[etf]

It is possible that I made some mistake while i transcripted solution here from my papers... Here is how i tested my solution in Matlab:
>> syms t C1 C2 C3
>> z=C1*exp(2*t)+C2*exp(t);
>> x=2*C1*exp(2*t)+C3*exp(-t);
>> y=diff(z,t)-x;
>> simplify(diff(x,t)-5*x-6*y+6*z)

ans =

0

>> simplify(diff(y,t)+2*x+3*y-4*z)

ans =

0

>> simplify(diff(z,t)-x-y)

ans =

0

 

[ehild]

It is possible that I made some mistake while i transcripted solution here from my papers... Here is how i tested my solution in Matlab:
>> syms t C1 C2 C3
>> z=C1*exp(2*t)+C2*exp(t);
>> x=2*C1*exp(2*t)+C3*exp(-t);
>> y=diff(z,t)-x;

So y(t)=C2 *exp(t)-C3*exp(-t).

To complete a solution, you have to give all functions : x(t), y(t), z(t).

Your solution is correct.

ehild