Question

Find the general solution of the differential equation y" + 16y = 2 sec(4t) tan(4t)

Select one:

O none of the given answers is true

y = ci cos(4t) + c2 sin(4t) + ,t cos(4t) - sin(4t) In | cos(4t) |

y = ci cos(4t) + c2 sin(4t) - t cos(4t) + sin(4t) In | cos(4t) |

y = ci cos(4t) + c2 sin (4t) -

7t sin(4t) - - cos(4t) In | sin(4t) |

y = ci cos(4t) + C2 sin(4t) + ,t sin(4t) + - cos(4t) In | sin(4t)|

Answer 1

Corresponding homogeneous differential equation

$$y''+16y=0$$

Characteristic equation

$$r^2+16=0$$

$$r=\pm4i$$

The general solution of the homogeneous differential equation is

$$y=C_1\cos(4t)+C_2\sin(4t)$$

Use the variation of parameters

$$W(y_1, y_2)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=\begin{vmatrix} \cos(4t) & \sin(4t) \\ -4\sin(4t) & 4\cos(4t) \end{vmatrix}=4$$

Find the partial solution

$$y_p=-\cos(4t)\int\dfrac{\sin(4t)\sec(4t)\tan(4t)}{4}dt$$

$$+\sin(4x)\int\dfrac{\cos(4x)\sec(x)\tan(x)}{4}dx$$

$$\int\dfrac{\sin(4t)\sec(4t)\tan(4t)}{4}dt=\int\dfrac{\tan^2(4t)}{4}dt$$

$$=\int\dfrac{\sec^2(4t)-1}{4}dt=\dfrac{\tan(4t)}{16}-\dfrac{t}{4}-C_3$$

$$\int\dfrac{\cos(4t)\sec(4t)\tan(4t)}{4}dt=\int\dfrac{\tan(4t)}{4}dt$$

$$=\int\dfrac{\sin(4t)}{4\cos(4t)}dt=-\dfrac{\ln|\cos(4t)|}{16}+C_4$$

$$y_p=-\cos(4t)\dfrac{\tan(4t)}{16}+\dfrac{t}{4}\cos(4t)-C_3\cos(4t)$$

$$-\sin(4t)\dfrac{\ln|\cos(4t)|}{16}+C_4\sin(4t)$$

The general solution of the given differential equation is

$$y=c_1\cos(4t)+c_2\sin(4t)$$

$$-\dfrac{\sin(4t)}{16}+\dfrac{t}{4}\cos(4t)-\sin(4t)\dfrac{\ln|\cos(4t)|}{16}$$