Question

(D^2-6D+13)y=8e^{3x}sin2x find CF and PI​

Answer 1

The given equation is,

$d^2y/dx^2 - 6 dy/dx +13y = 8 e^3x Sin 2x —(1)$

Let,

$y= e^mx$ be a trial solution of equation $d^2y/dx^2 - 6 dy/dx +13y = 0 — (2)$

• Now;

          $dy/dx = me^mx$

           and

          $d^2y/dx^2 = m^2 e^mx$

• Now lets Put these values in equation (2),

we get;

          $m^2 e^mx - 6me^mx + 13 e^mx = 0$

            $e^mx ( m^2 - 6m +13) = 0$

              $e^mx ≠ 0 so, m^2 - 6m +13 =0$

by solving this with the equation of sridhara charya we get,

$m = [- (-6)±Sqrt{(-6)^2 - 4×1×13}]/ 2×1$

$m = {6±Sqrt(-16)}/2$

$m = {6±Sqrt(-16)}/2$

$m = 3±2i , where i = sqrt(-1)$

• Now;

          the Complementary function of the given equation (1) is ;

                $C. F. = e^3x( A Cos 2x + B Sin 2x)$

• Now;

we found the value of a particular integral with help of D -the operator's method ;

                  P$. I. = 8 e^3x Sin 2x / (D^2 - 6D+13)$

                        $= 8 e^3x [Sin 2x / {(D+3)^2 - 6(D+3) +13}]$

                        $= 8 e^3x [ Sin 2x / (D^2 + 4)]$

                         $= 8 x e^3x {Sin 2x / 2D}$

                       $= 4 x e^3x (Sin 2x /D)$

                      $= 4 x e^3x (- Cos 2x/2)$

                     $= - 2 x e^3x Cos 2x$

• Therefore; the required general solution of given equation(1) is,

                  $y = C. F. + P. I.$

                $y = e^3x (A Cos 2x + B Sin 2x) - 2 x e^3x Cos 2x$

( where A&B are the arbitrary constants)

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