Answer to Question #163041 in Differential Equations for Leyvanhong

Question

dx/dt=2x+2y, dy/dt=x+3y

Answer 1

Want to solve the following system:,

$\frac{dx}{dt}=2x+2y\hspace{10pt}(1)\\ \frac{dy}{dt}=x+3y\hspace{15pt}(2)$

i.e., want to find $x(t)$ and $y(t)$ which satisfy the above two equations.

Differentiating equation (1) with respect to $t$ yields,

$\frac{d^2 x}{dt^2}=2\frac{dx}{dt}+2\frac{dy}{dt}$

Substituting $\frac{dy}{dt}$ from equation (2), we obtain,

$\frac{d^2x}{dt^2}=2\frac{dx}{dt}+(2x+6y)$

Substituting $y$ from equation (1) gives,

$\frac{d^2x}{dt^2}=2\frac{dx}{dt}+2x+3\left(\frac{dx}{dt}-2x\right)\\ \Rightarrow \frac{d^2x}{dt^2}=5\frac{dx}{dt}-4x\\ \Rightarrow \frac{d^2x}{dt^2}-5\frac{dx}{dt}+4x = 0\hspace{20pt}(3)$

Clearly equation (3) has a solution of the form,

$x(t)=Ae^{r_1 t}+Be^{r_2 t}$ , where $A, B$ are two fixed constants and $r_1$ and $r_2$ are the two roots of the equation:

$r^2 - 5r+4=0\\ \Rightarrow (r-1)(r-4)=0\\ \Rightarrow r=1\text{ or }r=4$

Thus we can take $r_1=1$ and $r_2=4$

Hence, $x(t)=Ae^t + Be^{4t}$

Using this value of $x(t)$ , we get $y(t)$ (from equation (1)) as follows,

$y(t)=\frac{1}{2}\left(\frac{dx}{dt}\right)-x\\ \Rightarrow y(t)=-\frac{A}{2}e^t + B e^{4t}$

So the solution of the system is given by,

$x(t)=Ae^t + Be^{4t}\\ y(t)=-\frac{A}{2}e^t + B e^{4t}$

The values of $A$ and $B$ has to be determined from the initial values, if provided (for example, those can be obtained if $x(0)$ and $y(0)$ are known.)