Partially Decoupled System with 3 variables

[Omzyma]

Hello!

I have the following initial value problem:
\[ x' = x + 2y + 3z \]
\[ y' = 4y + 5z \]
\[ z' = 6z \]

All I'm looking to do is find the general solution to this system, and as long as I'm doing this correctly I have these answers:

\[ y(t) = K_2e^{4t} + \tfrac{5K_1}{2}e^{6t} \]
\[ z(t)=K_1e^{6t} \]

But I'm having trouble getting x(t) to finish this up. Any help on what my next steps should be?

 

[topsquark]

Hello!

I have the following initial value problem:
\[ x' = x + 2y + 3z \]
\[ y' = 4y + 5z \]
\[ z' = 6z \]

All I'm looking to do is find the general solution to this system, and as long as I'm doing this correctly I have these answers:

\[ y(t) = K_2e^{4t} + \tfrac{5K_1}{2}e^{6t} \]
\[ z(t)=K_1e^{6t} \]

But I'm having trouble getting x(t) to finish this up. Any help on what my next steps should be?

I agree with your solutions to z(t) and y(t). What's the difficulty? You will have
\(\displaystyle x' = x + 2 \left ( K_2 e^{4t} \right ) + 3 \left ( \dfrac{5}{2} K_1 e^{6t} \right )\)

You've shown you can solve the homogeneous equation. So pick your particular solution: \(\displaystyle x_p = A \left ( e^{4t} \right )+ B \left ( e^{6t} \right )\)

-Dan

 

[Omzyma]

I agree with your solutions to z(t) and y(t). What's the difficulty? You will have
\(\displaystyle x' = x + 2 \left ( K_2 e^{4t} \right ) + 3 \left ( \dfrac{5}{2} K_1 e^{6t} \right )\)

You've shown you can solve the homogeneous equation. So pick your particular solution: \(\displaystyle x_p = A \left ( e^{4t} \right )+ B \left ( e^{6t} \right )\)

-Dan

Thanks for your reply!

Picking \(\displaystyle x_p = A \left ( e^{4t} \right )+ B \left ( e^{6t} \right )\) as my particular solution I ended up with \(\displaystyle 3A \left ( e^{4t} \right )+ 5B \left ( e^{6t} \right ) = 2K_2 \left ( e^{4t} \right ) + 3K_1 \left ( e^{6t} \right )\)

At this point do I solve for both A and B simultaneously, making either A or B as function of the other? Or should I just make like terms equal to each other and solve for A and B individually?

 

[topsquark]

Thanks for your reply!

Picking \(\displaystyle x_p = A \left ( e^{4t} \right )+ B \left ( e^{6t} \right )\) as my particular solution I ended up with \(\displaystyle 3A \left ( e^{4t} \right )+ 5B \left ( e^{6t} \right ) = 2K_2 \left ( e^{4t} \right ) + 3K_1 \left ( e^{6t} \right )\)

At this point do I solve for both A and B simultaneously, making either A or B as function of the other? Or should I just make like terms equal to each other and solve for A and B individually?

If you have \(\displaystyle 3A \left ( e^{4t} \right )+ 5B \left ( e^{6t} \right ) = 2K_2 \left ( e^{4t} \right ) + 3K_1 \left ( e^{6t} \right )\) true for all t then \(\displaystyle 3A = 2 K_2\) and \(\displaystyle 5B = 3 K_1\).

-Dan

 

[Omzyma]

If you have \(\displaystyle 3A \left ( e^{4t} \right )+ 5B \left ( e^{6t} \right ) = 2K_2 \left ( e^{4t} \right ) + 3K_1 \left ( e^{6t} \right )\) true for all t then \(\displaystyle 3A = 2 K_2\) and \(\displaystyle 5B = 3 K_1\).

-Dan

So that would mean \(\displaystyle A=\tfrac{2}{3} \) and \(\displaystyle B = \tfrac{3}{5}! \)

Thanks for your help. Just needed a little nudge I guess.