How Do You Solve a Partially Decoupled System of Differential Equations?

[superduck1385]

Consider the partially-decoupled system
dx/dt = x+2y+1
dy/dt = 3y

Derive the general solution. Find equilibrium points. Find the solution satisfying the initial value (x(sub 0), y(sub 0)) = (-1,3).

 

[MarkFL]

Hello and welcome to MHB, superduck1385! :D

We normally ask that people posting questions show what they have done so far so that our helpers know exactly where you are stuck or may be going wrong, and can offer the best help possible.

Now, I think my first step would be to solve the second equation, as it is linear. Perhaps the simplest method would be to observe that it has one characteristic root...can you find this root, and thus the general solution to the second equation, giving you something which you can substitute into the first equation?

 

[MarkFL]

Since more than 48 hours has gone by with no additional feedback from the OP, I will finish the problem for the benefit of future readers.

Now, we see that the second equation has the characteristic root:

\(\displaystyle r=3\)

and so the general solution for that equation is:

\(\displaystyle y(t)=c_1e^{3t}\)

Now, substituting this into the first equation, there results:

\(\displaystyle \d{x}{t}=x+2c_1e^{3t}+1\)

Writing this ODE in standard linear form, we obtain:

\(\displaystyle \d{x}{t}-x=2c_1e^{3t}+1\)

We see by inspection that the integrating factor is:

\(\displaystyle \mu(t)=e^{-t}\)

And hence, then ODE becomes:

\(\displaystyle e^{-t}\d{x}{t}-xe^{-t}=2c_1e^{2t}+e^{-t}\)

We may now rewrite the left side:

\(\displaystyle \frac{d}{dt}\left(e^{-t}x\right)=2c_1e^{2t}+e^{-t}\)

Integrating both sides with respect to $t$, we obtain:

\(\displaystyle e^{-t}x=c_1e^{2t}-e^{-t}+c_2\)

And so we obtain:

\(\displaystyle x(t)=c_1e^{3t}+c_2e^{t}-1\)

The general solution is thus:

\(\displaystyle Y(t)=\left(x(t),y(t)\right)=\left(c_1e^{3t}+c_2e^{t}-1,c_1e^{3t}\right)\)

To find the equilibrium points, we equate both derivatives to zero:

\(\displaystyle x+2y+1=0\)

\(\displaystyle 3y=0\)

The second equation implies $y=0$ and so the first equation implies $x=-1$, and so the sole equilibrium point is:

\(\displaystyle (x,y)=(-1,0)\)

Using the conventional notation:

\(\displaystyle x_0=x(0)\) and \(\displaystyle y_0=y(0)\)

we may determine the particular solution satisfying the given initial conditions as follows:

\(\displaystyle x(0)=c_1+c_2-1=-1\implies c_1+c_2=0\)

\(\displaystyle y(0)=c_1=3\implies c_2=-3\)

Hence, we find the particular solution:

\(\displaystyle Y_p(t)=\left(x(t),y(t)\right)=\left(3e^{3t}-3e^{t}-1,3e^{3t}\right)\)