Differential Equations System Solutions

  • #1
dmoney123
32
1

Homework Statement



Consider the initial value problem for the system of first-order differential equations

y_1' = -2y_2+1, y_1(0)=2

y_2' = -8y_1+2, y_2(0)=-1

If the matrix

[ 0 -2
-8 0 ]

has eigenvalues and eigenvectors L_1= -4 V_1= [ 1
2 ]

L_2=4 V_2= [ 1
-2]

then its solution will be:


Homework Equations




The Attempt at a Solution


[/B]
e^(-4t) +e^ (4t) from eigenvalues

multiply by respective eigenvectors and set to initial conditions gives 2 sets of equations and two unknown coefficients

2=c_1*e^(-4t)+c_2*e^(4t)
-1=c_1*2e^(-4t)+c_2*-2e^(4t)

c_1=3/4

c_2=5/4

I am very confident with these values being right for the coefficients, I know need to know how to use these to form a general solution.

I plug these values back into get

y_1=3/4e^(-4t)+5/4e^(4t)
y_2=3/2e^(-4t)-5/2e^(4t)



Then solution given is

y_1(t)=5/4e^(4t)+1/2e^(-4t)+1/4,
y_2(t)=-5/2e^(4t)+e^(-4t)+1/2

any help is appreciated! thanks!
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,915

Homework Statement



Consider the initial value problem for the system of first-order differential equations

y_1' = -2y_2+1, y_1(0)=2

y_2' = -8y_1+2, y_2(0)=-1

If the matrix

[ 0 -2
-8 0 ]

has eigenvalues and eigenvectors L_1= -4 V_1= [ 1
2 ]

L_2=4 V_2= [ 1
-2]

then its solution will be:


Homework Equations




The Attempt at a Solution


[/B]
e^(-4t) +e^ (4t) from eigenvalues
This is the solution of the homogeneous part only. The system is inhomogeneous, you have to add the particular solution of the equations to the general solution of the homogeneous part.
 
  • #3
dmoney123
32
1
This is the solution of the homogeneous part only. The system is inhomogeneous, you have to add the particular solution of the equations to the general solution of the homogeneous part.

I have not encountered this type of problem before. How do you go about finding the particular solution?
 
  • #4
ehild
Homework Helper
15,543
1,915
I have not encountered this type of problem before. How do you go about finding the particular solution?
It is simple in this case. Set both y1'=0 and y2' =0. What are y1 and y2 then?
 
  • #5
dmoney123
32
1
It is simple in this case. Set both y1'=0 and y2' =0. What are y1 and y2 then?

Okay, I work this out to y_1=1/4 and y_2=1/2...

which seems to fit in the equation... but I am still not really sure why the coefficients for the e^(-4t) don't seem to match the solution
 
  • #6
ehild
Homework Helper
15,543
1,915
Okay, I work this out to y_1=1/4 and y_2=1/2...

which seems to fit in the equation... but I am still not really sure why the coefficients for the e^(-4t) don't seem to match the solution
The general solution is
y1=c1e-4t+c2e4t+1/4
y2=c1e-4t+c2e4t+1/2

Fit these equations to the initial conditions y1(0)=2, y2(0)=-1.
 
  • #7
dmoney123
32
1
The general solution is
y1=c1e-4t+c2e4t+1/4
y2=c1e-4t+c2e4t+1/2

Fit these equations to the initial conditions y1(0)=2, y2(0)=-1.

Thank you so much! That makes total sense!
 
  • #8
ehild
Homework Helper
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You are welcome. :)
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
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970
iHere is how I would have done this, without using "matrices": differentiate the first equation again to get [itex]y_1''= -2y_2'[/itex]. But from the second equation, [itex]y_2'= -8y_1+ 2[/itex] so that [itex]y_1''= -2(-8y_1+ 2)= 16y_1- 4[/itex] or [itex]y_1''- 16y_1= -4[/itex]. The associated homogeneous equation, [itex]y_1''- 16y= 0[/itex] has characteristic equation [itex]r^2- 16= 0[/itex] which has roots [itex]r= 4[/itex] and [itex]r= -4[/itex] so the general solution to the associated homogeneous equation is [itex]y(t)= C_1e^{4t}+ C_2e^{-4t}[/itex].

Now new look for a particular solution to the entire equation. Since the right hand, "non-homogeneous", part is the constant -4, we try a solution of the form y= A for A some constant. Then y'= y''= 0 so the equation becomes [itex]0- 16A= -4[/itex] and [itex]A= 1/4[/itex].

That is, [itex]y_1(t)= C_1e^{4t}+ C_2e^{-4t}+ \frac{1}{4}[/itex]. Now, we can write the first equation, [itex]y_1'= -2y_2+ 1[/itex] is the same as [itex]2y_2= -y_1'+ 1[/itex] or [itex]y_1= -(1/2)y_2'+ 1/2[/itex]. Since [itex]y_1(t)= C_1e^4t+ C_2e^{-4t}+ \frac{1}{4}[/itex], [itex]y_1'= 4C_1e^{4t}- 4C_2e^{-4t}[/itex], [itex]-(1/2)y_2'= -2C_1e^{4t}+ 2e^{-4t}[/itex] and [itex]y_1(t)= -2C_1e^{4t}+ 2C_2e^{-4t}+ 1/2[/itex].
 
Last edited by a moderator:
  • #10
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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Homework Statement



Consider the initial value problem for the system of first-order differential equations

y_1' = -2y_2+1, y_1(0)=2

y_2' = -8y_1+2, y_2(0)=-1

If the matrix

[ 0 -2
-8 0 ]

has eigenvalues and eigenvectors L_1= -4 V_1= [ 1
2 ]

L_2=4 V_2= [ 1
-2]

then its solution will be:


Homework Equations




The Attempt at a Solution


[/B]
e^(-4t) +e^ (4t) from eigenvalues

multiply by respective eigenvectors and set to initial conditions gives 2 sets of equations and two unknown coefficients

2=c_1*e^(-4t)+c_2*e^(4t)
-1=c_1*2e^(-4t)+c_2*-2e^(4t)

c_1=3/4

c_2=5/4

I am very confident with these values being right for the coefficients, I know need to know how to use these to form a general solution.

I plug these values back into get

y_1=3/4e^(-4t)+5/4e^(4t)
y_2=3/2e^(-4t)-5/2e^(4t)



Then solution given is

y_1(t)=5/4e^(4t)+1/2e^(-4t)+1/4,
y_2(t)=-5/2e^(4t)+e^(-4t)+1/2

any help is appreciated! thanks!

Re-write your system as
[tex] y_1' = -2y_2 + 1 = -2\left(y_2 - \frac{1}{2} \right) \\
y_2' = -8y_1+2 = -8\left(y_1 - \frac{1}{4} \right) [/tex]
Since ##d(y_1 - 1/4)/dt = dy_1/dt## and ##d(y_2 - 1/2)/dt = dy_2/dt##, the variables ##Y_1 = y_1 - 1/4## and ##Y_2 = y_2 - 1/2)## satisfy the homogeneous system
[tex] Y_1' = -2Y_2\\
Y_2' = -8 Y_1
[/tex]
You can easily figure out the boundary values ##Y_1(0), \, Y_2(0)##.

This type of trick always works whenever the non-homogeneous terms on the right are constants.
 

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