Question

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

Answer 1

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}$

Step-by-step explanation:

Given differential equation is

   y"+y'-2y = 1

$=>\ (D^2\ +\ D\ -\ 2)y\ =\ 1$

Hence, the characteristics is

   $D^2+D-2\ =\ 0$

$=>\ D^2\ +\ 2D\ -\ D\ -2\ =\ 0$

$=> (D+2)(D-1) = 0$

=> D = -2, 1

The general equation of the given differential equation is

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's consider that

$y_1(t)\ =\ e^{-2t}$              $y_2(t)\ =\ e^{t}$

$y'_1(t)\ =\ -2e^{-2t}$          $y'_2(t)\ =\ e^t$

g(t) = 1

Wronskian can be given by,

W = y_1(t)y'_2(t) - y_2(t)y'_1(t)

   $=\ e^{-2t}.e^t\ -\ e^{t}.(-2e^{-2t})$

   $=\ e^{-t}\ +\ 2e^{-t}$

   $=\ 3e^{-t}$

Now, the particular integral can be given by

$y_p(t)\ = \ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}\ +\ y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

      $=\ -e^{-2t}\int{\dfrac{e^t\times 1}{3e^{-t}}dt}\ +\ e^t\int{\dfrac{e^{-2t}\times 1}{3e^{-t}}dt}$

      $=\ -e^{-2t}\int{\dfrac{e^{2t}}{3}dt}\ +\ e^t\int{\dfrac{e^{-t}}{3}dt}$

      $=\ (-e^{-2t})(\dfrac{e^{2t}}{6})\ +\ (e^t)(\dfrac{e^{-t}}{-3})$

      $=\ \dfrac{-1}{6}\ -\ \dfrac{1}{3}$

      $=\ \dfrac{-3}{6}$

      $=\ \dfrac{-1}{2}$

Now,

$y(t)\ =\ y_c(t)\ +\ y_p(t)$

      $=\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}$

Hence, the complete solution of the given differential equation is

$y(t)\ =\ C_1e^{-2t}+C_2e^t\ -\ \dfrac{1}{2}$