Answer to Question #257263 in Differential Equations for ron

Question

solve the system of first-order linear differential equations

y1'=y1-y2

y'2=2y1+4y2

Answer 1

Let us solve the system of first-order linear differential equations

$\begin{cases} y_1'=y_1-y_2\\ y_2'=2y_1+4y_2 \end{cases}$

It follows that $y_2=y_1-y_1',$ and hence $y_2'=y_1'-y_1''.$ Consequently, we get the equation

$y_1'-y_1''=2y_1+4(y_1-y_1'),$ which is equivalent to $y_1''-5y_1'+6y_1=0.$

The characteristic equation $k^2-5k+6=0$ of $y_1''-5y_1'+6y_1=0$ is equivalent to $(k-2)(k-3)=0,$ and thus it has the roots $k_1=2,\ k_2=3.$ Therefore the solution of this differential equation is $y_1=C_1e^{2x}+C_2e^{3x}.$

Then $y_1'=2C_1e^{2x}+3C_2e^{3x},$ and therefore

$y_2=y_1-y_1'=C_1e^{2x}+C_2e^{3x}-(2C_1e^{2x}+3C_2e^{3x})=-C_1e^{2x}-2C_2e^{3x}.$

We conclude that the general solution of the system is of the form:

$\begin{cases} y_1=C_1e^{2x}+C_2e^{3x}\\ y_2=-C_1e^{2x}-2C_2e^{3x} \end{cases}.$