Question

Solve the below differential equation
y'''+3y''+2y'-5y=sin2t

Answer 1

Consider the equation

$$y'''+3y''+2y'-5y=0$$

We write the characteristic equation

$$k^3+3\times k^2+2\times k -5=0$$

Using the Cardano formulas we get the roots of the equation

$$k_1=0.9$$

$$k_2=-1.95-1.31\times i$$

$$k_3= -1.95 +1.31\times i$$

where

$$i=\sqrt {-1}.$$

Therefore solving the equation

$$y'''+3y''+2y'-5y=0$$

$$y=C_1\times e^ {0.9t} + C_2\times e^ {-1.95t}sin (1.31t)+C_3\times e^ {-1.95t}cos (1.31t)$$

where

$$C_1, C_2, C_3$$

are constants. Assume a partial solution of the equation is

$$y_p=Asin(2t)+Bcos(2t)$$

where A, B are constants. Hence

$$y'_p=2Acos(2t)-2Bsin(2t)$$

$$y''_p=-4Asin(2t)-4Bcos(2t)$$

$$y'''_p=-8Acos(2t)+8Bsin(2t)$$

Substituting in the equation we get

$$(-3A+6B)sin(2t)+(-6A-3B)cos(2t)=sin(2t).$$

As result one gets

$$A=- \frac {1} {15}$$

$$B= \frac{2} {15}$$

Therefore the solution of the equation

$$y'''+3y''+2y'-5y=sin(2t)$$

is

$$y=C_1\times e^ {0.9t} + C_2\times e^ {-1.95t}sin (1.31t)+C_3\times e^ {-1.95t}cos (1.31t)+$$

$$- \frac {1} {15} sin(2t) + \frac {2} {15} cos(t)$$

where

$$C_1, C_2, C_3$$

are constants.

Answer.

$$y=C_1\times e^ {0.9t} + C_2\times e^ {-1.95t}sin (1.31t)+C_3\times e^ {-1.95t}cos (1.31t)- \frac {1} {15} sin(2t) + \frac {2} {15} cos(t)$$

where

$$C_1, C_2, C_3$$

are constant.