Question

Obtain the differential equation from the given general solution.

y= c1 cos(4x) +c2 sin(4x) +c3 xcos4x +c4x sin(4x)

Answer 1

The steps necessary to find the ordinary differential equations satisfied by this solution are

Differentiate the general solution with respect to $x$ exactly $4$  times.

Use the $(4+1)$ number of expressions ($4$ derivatives) obtained to eliminate the $4$  arbitrary constants in terms of the dependent variable or its derivatives.

Obtain the final expression which contains absolutely no arbitrary constant. This is the required differential equation.

$$y= c_1 \cos(4x) +c_2 \sin(4x) +c_3 x\cos(4x)$$

$$+c_4x \sin(4x)$$

$$y'=-4c_1 \sin(4x) +4c_2 \cos(4x) -4c_3 x\sin4x$$

$$+c_3 \cos(4x)+4c_4x\cos(4x)+c_4\sin(4x)$$

$$y''= -16c_1 \cos(4x) -16c_2 \sin(4x) -16c_3 x\cos(4x)$$

$$-8c_3\sin(4x)-16c_4x\sin(4x)+8c_4\cos(4x)$$

$$y'''= 64c_1 \sin(4x) -64c_2 \cos(4x)+64c_3 x\sin(4x)$$

$$-48c_3\cos(4x)-64c_4x\cos(4x)-48c_4\sin(4x)$$

$$y^{''''}= 256c_1 \cos(4x)+256c_2 \sin(4x) +256c_3 x\cos(4x)$$

$$+256c_3\sin(4x)+256c_4x\sin(4x)-256c_4\cos(4x)$$

$$y^{''''}+16y^{''}=128c_3\sin(4x)-128c_4\cos(4x)$$

$$y^{'''}+16y^{'}=-32c_3\cos(4x)-32c_4\sin(4x)$$

$$y^{''}+16y= -8c_3\sin(4x)+8c_4\cos(4x)$$

$$y^{''''}+32y^{''}+256y=0$$

The differential equation from the given general solution

$$y^{''''}+32y^{''}+256y=0$$

Auxiliary (corresponding) equation

$$r^2+32r^2+256=0$$

$$(r^2+16)^2=0$$

$$r_1=r_2=-4i, r_3=r_4=4i$$