Q:

Use the variation of parameters method to solve the DE y"+y'- 2y=1

Accepted Solution

A:
Answer:[tex]y(t)\ =\ C_1.e^{-2t}+C_2e^t-\ t.\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]Step-by-step explanation:Given differential equation is,      y"+y'-2y=1[tex]=>\ (D^2+D-2D)y\ =\ 1[/tex]To find the complementary function we will write,     [tex]D^2+D-2=0[/tex][tex]=>\ D\ =\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}[/tex][tex]=>\ D\ =\ -2\ or\ 1[/tex]Hence, the complementary function can be given by[tex]y(t)\ =\ C_1e^{-2t}\ +\ C_2e^t[/tex]Let's say,[tex]y_1(t)\ =\ e^{-2t}\ \ =>y'_1(t)\ =\ -2e^{-2t}[/tex][tex]y_2(t)\ =\ e^{t}\ \ =>y'_2(t)\ =\ e^{t}[/tex][tex]g(t)\ =\ 1[/tex]Wronskian can be given by,[tex]W\ =\ y_1(t).y'_2(t)\ -\ y_2(t).y'_1(t)[/tex]      [tex]=\ e^{-2t}.e^{t}\ -\ e^{t}.(-2e^{-2t})[/tex]      [tex]=\ e^{-t}\ +\ 2e^{-t}[/tex]      [tex]=\ 3.e^{-t}[/tex]Now, the particular integral can be given by[tex]y_p(t)=\ -y_1(t)\int\dfrac{y_2(t).g(t)}{W}dt\ +\  y_2(t)\int\dfrac{y_1(t).g(t)}{W}dt[/tex]         [tex]=\ -e^{-2t}\int\dfrac{e^t.1}{3.e^{-t}}+e^t\int\dfrac{e^{-2t}.1}{3.e^{-t}}dt[/tex]        [tex]=\ -e^{-2t}\int\dfrac{1}{3}dt+\dfrac{e^t}{3}\int e^{-t}dt[/tex]        [tex]=\ \dfrac{-e^{-2t}}{3}.t\ -\ \dfrac{e^t}{3}.e^{-t}[/tex]         [tex]=\ -t.\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]Hence, the complete solution can be given by[tex]y(t)\ =\ C_1.e^{-2t}+C_2e^t-\ t.\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]