# Differential equation is hard

!!!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!

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...

Last edited:

## Answers and Replies

Related Calculus and Beyond Homework Help News on Phys.org
HallsofIvy
Science Advisor
Homework Helper
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
$$k_1= \frac{2+ 4i}{4}k_2$$

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

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:
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

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

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

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

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

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!

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

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

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...?

Do you still not see the glaring algebra error in the bottom left...?
Thank you. That's all I needed....

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