Answer to Question #330181 in Differential Equations for mnamed

Question

solve system of linear differential equations  dx/dt-4y=1 and dy/dt+x=2

Answer 1

From second equation:

$x = 2 - \frac{{dy}}{{dt}} \Rightarrow \frac{{dx}}{{dt}} = - \frac{{{d^2}y}}{{d{t^2}}}$

Substitute into the first equation:

$- \frac{{{d^2}y}}{{d{t^2}}} - 4y = 1 \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} + 4y = - 1$

We solve the resulting differential equation.

Characteristic equation:

${k^2} + 4 = 0 \Rightarrow {k^2} = - 4 \Rightarrow k = \pm 2i$

Then the general solution of the homogeneous equation has the form:

${y_0} = {C_1}\cos 2t + {C_2}\sin 2t$

A particular solution will be sought in the form:

$Y = A \Rightarrow \frac{{dY}}{{dt}} = \frac{{{d^2}Y}}{{d{t^2}}} = 0$

Substitute into the equation:

$4A = - 1 \Rightarrow A = Y = - \frac{1}{4}$

Then

$y = {y_0} + Y = {C_1}\cos 2t + {C_2}\sin 2t - \frac{1}{4} \Rightarrow \frac{{dy}}{{dt}} = - 2{C_1}\sin 2t + 2{C_2}\cos 2t$

$x = 2 - \frac{{dy}}{{dt}} = 2 + 2{C_1}\sin 2t - 2{C_2}\cos 2t$

Answer: $y = {C_1}\cos 2t + {C_2}\sin 2t - \frac{1}{4},\,\,x = 2 + 2{C_1}\sin 2t - 2{C_2}\cos 2t$