Question

solve dx/df=3x-4y, dy/dt=x-y

Answer 1

From the second equation we get:

$x = \frac{{dy}}{{dt}} + y = y' + y \Rightarrow \frac{{dx}}{{dt}} = x' = y'' + y'$

Substitute the found values ​​into the first equation:

$y'' + y' = 3\left( {y' + y} \right) - 4y \Rightarrow y'' - 2y' + y = 0$

Characteristic equation:

${k^2} - 2k + 1 = 0 \Rightarrow {(k - 1)^2} = 0 \Rightarrow {k_1} = {k_2} = 1$

Then

$y = {C_1}{e^t} + t{C_2}{e^t} \Rightarrow y' = {C_1}{e^t} + {C_2}{e^t} + t{C_2}{e^t}\Rightarrow \\ \Rightarrow x = y' + y = {C_1}{e^t} + {C_2}{e^t} + t{C_2}{e^t} + {C_1}{e^t} + t{C_2}{e^t} = \\=2{C_1}{e^t} + 2t{C_2}{e^t} + {C_2}{e^t}$

Answer: $y = {C_1}{e^t} + t{C_2}{e^t},\,x = 2{C_1}{e^t} + 2t{C_2}{e^t} + {C_2}{e^t}$