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Differential equation is hard

2,981
2
!!!For The Love of GOD!!! Diff Eq

1. Homework Statement

I have done this problem 3 times. I am getting a sign error somewhere and I cannot find it.

When I solve this DE and then plug it back into the original, it is not checking out!!! Can anybody see my error? I would really appreciate a second set of eyes here. Clearly I am at that point where I just keep seeing what I think I am supposed to see!



Picture1-6.png




Re writing the solution to the DE with the new constants is

[tex]x=-3/2\cos4t+3\sin4t+7/2\cos4t+1/2\sin4t=2\cos4t+19/4\sin4t[/tex]
[tex]y=3\cos4t-7/2\sin4t[/tex]

But when I differentiate and plug back into either if the originals, I am coming up with two sign errors...
 
Last edited:

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,732
893
At one point in your solution you have:
[tex]4 k_1+ (-2- 4i)k_2= 0[/tex]
and then
[tex]k_1= \frac{-2- 4i}{4}k_2[/tex]
It should be
[tex]k_1= \frac{2+ 4i}{4}k_2[/tex]
 
2,981
2
At one point in your solution you have:
[tex]4 k_1+ (-2- 4i)k_2= 0[/tex]
and then
[tex]k_1= \frac{-2- 4i}{4}k_2[/tex]
It should be
[tex]k_1= \frac{2+ 4i}{4}k_2[/tex]
I do not understand why it should be positive? From the matrix I have [itex]4k_1+(-2-\lambda)k_2[/itex] where lambda=4i......unless I am interpreting matrices wrong??
 
Vid
401
0
Basic Algebra.

4K1 - (2+4i)k2 = 0
K1 = (2+4i)k2/4
 
2,981
2
Basic Algebra.

4K1 - (2+4i)k2 = 0
K1 = (2+4i)k2/4
Okay. Obviously I am misinterpreting how a matrix operates. Not how to do basic algebra.






























I was just kidding, maybe that was inappropriate...my bad.
 
Last edited:
2,981
2
If the original equation was
x'=2x-5y
y'=4x-2y

To find the charectaristic equation I create this matrix thing...I don't know what it is called

[tex]
\left[\begin{array}{cc}2-\lambda & -5 \\ 4 & -2-\lambda\end{array}\right]=0[/tex]

Now if I plug in [itex]\lambda= 4i[/itex] to the 2nd row, how do you suppose I will get anything positive?
 
2,981
2
If the original equation was
x'=2x-5y
y'=4x-2y

To find the charectaristic equation I create this matrix thing...I don't know what it is called

[tex]
\left[\begin{array}{cc}2-\lambda & -5 \\ 4 & -2-\lambda\end{array}\right]=0[/tex]

Now if I plug in [itex]\lambda= 4i[/itex] to the 2nd row, how do you suppose I will get anything positive?
Can anybody see what I am doing wrong in this procedure? What am I misunderstanding?

If I plug [itex]\lambda=4i[/itex] into [itex]-2-\lambda[/itex] why would I get 2+4i ?

I have five more problems similar to this and cannot move on until I clear this up.

Any help would be great!
 
2,981
2
If the original equation was
x'=2x-5y
y'=4x-2y

To find the charectaristic equation I create this matrix thing...I don't know what it is called

[tex]
\left[\begin{array}{cc}2-\lambda & -5 \\ 4 & -2-\lambda\end{array}\right]=0[/tex]

Now if I plug in [itex]\lambda= 4i[/itex] to the 2nd row, how do you suppose I will get anything positive?
Can anybody see what I am doing wrong in this procedure? What am I misunderstanding?

If I plug [itex]\lambda=4i[/itex] into [itex]-2-\lambda[/itex] why would I get 2+4i ?

I have five more problems similar to this and cannot move on until I clear this up.

Any help would be great!

Kind of desperate here.....
 
Vid
401
0
Do you still not see the glaring algebra error in the bottom left...?
 
2,981
2
Do you still not see the glaring algebra error in the bottom left...?
Thank you. That's all I needed....
 
308
0
Your K1 value is where the sign error is. Change it to what Hallsofivy did and solve again from there.
 

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