# Partially Decoupled System with 3 variables

• MHB
• Omzyma
In summary, this individual is having trouble getting x(t) to finish solving the system and is looking for guidance on what to do next.f

#### 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?

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

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

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?

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

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.