Diff Eq Sign Error - Seeking Second Set of Eyes

[Saladsamurai]

!For The Love of GOD! Diff Eq

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!

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

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

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

 

[HallsofIvy]

At one point in your solution you have:
4 k_1+ (-2- 4i)k_2= 0
and then
k_1= \frac{-2- 4i}{4}k_2
It should be
k_1= \frac{2+ 4i}{4}k_2

 

[Saladsamurai]

At one point in your solution you have:
4 k_1+ (-2- 4i)k_2= 0
and then
k_1= \frac{-2- 4i}{4}k_2
It should be
k_1= \frac{2+ 4i}{4}k_2

I do not understand why it should be positive? From the matrix I have 4k_1+(-2-\lambda)k_2 where lambda=4i...unless I am interpreting matrices wrong??

 

[Vid]

Basic Algebra.

4K1 - (2+4i)k2 = 0
K1 = (2+4i)k2/4

 

[Saladsamurai]

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.

 

[Saladsamurai]

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

<br /> \left[\begin{array}{cc}2-\lambda &amp; -5 \\ 4 &amp; -2-\lambda\end{array}\right]=0

Now if I plug in \lambda= 4i to the 2nd row, how do you suppose I will get anything positive?

 

[Saladsamurai]

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

<br /> \left[\begin{array}{cc}2-\lambda &amp; -5 \\ 4 &amp; -2-\lambda\end{array}\right]=0

Now if I plug in \lambda= 4i 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 \lambda=4i into -2-\lambda 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!

 

[Saladsamurai]

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

<br /> \left[\begin{array}{cc}2-\lambda &amp; -5 \\ 4 &amp; -2-\lambda\end{array}\right]=0

Now if I plug in \lambda= 4i 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 \lambda=4i into -2-\lambda 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]

Do you still not see the glaring algebra error in the bottom left...?

 

[Saladsamurai]

Do you still not see the glaring algebra error in the bottom left...?

Thank you. That's all I needed...

 

[happyg1]

Your K1 value is where the sign error is. Change it to what Hallsofivy did and solve again from there.