Answer to Question #161123 in Differential Equations for H BUDELI

Question

Find the solution of the following couple of different differential equations x^'=4x-y and y^'=-4x+4y when t=0 ,x(0)=1 and y(0)=1

Answer 1

$\begin{cases} x^\prime =4x-y \\ y^\prime=-4x+4y \end{cases}$ and $x(0)=y(0)=1$ .

From the first equation we get $x^{\prime \prime}=4x^\prime -y^\prime =4x^\prime -\left(-4x+4y \right)=4x^\prime +4x-4y=4x^\prime +4x-4(4x-x^\prime )=4x^\prime +4x-16x+4x^\prime$

We have second order differential equation $x^{\prime \prime}-8x^\prime +12x=0$ .

The characteristic equation is $\lambda ^2-8\lambda +12=0$ .

The roots are $\lambda _1=2,\$ $\lambda _2=6$ .

The general solution is $x(t)=c_1e^{2t}+c_2e^{6t}$ .

Then we can find $y=4x-x^\prime =4\left( c_1e^{2t}+c_2e^{6t} \right) -\left(2c_1e^{2t}+6c_2e^{6t}\right)= 2c_1e^{2t}-2c_2e^{6t}$ .

Using $x(0)=y(0)=1$ , we get a system:

$\begin{cases} 1=c_1+c_2 \\ 1=2c_1-2c_2 \end{cases}$ $\begin{cases} 2=2c_1+2c_2 \\ 1=2c_1-2c_2 \end{cases}$ $\begin{cases} c_1=\tfrac{3}{4} \\ c_2=\tfrac{1}{4} \end{cases}$

Answer: $x(t)=\tfrac{1}{4}\left(3e^{2t}+e^{6t}\right), \ \ y(t)=\tfrac{1}{2}\left(3e^{2t}-e^{6t}\right)$ .