Question

(D⁴ - 64) y= x cos x

Answer 1

$\left( {{D^4} - 64} \right)y = x\cos x \Rightarrow {y^{(4)}} - 64y = x\cos x$

The characteristic equation has the form

${k^4} - 64 = 0$

$({k^2} - 8)({k^2} + 8) = 0$

$(k - \sqrt 8 )(k + \sqrt 8 )(k - \sqrt 8 i)(k + \sqrt 8 i) = 0$

${k_{1,2}} = \pm \sqrt 8 ,\,\,{k_{3,4}} = \pm \sqrt 8 i$

Then the general solution of the homogeneous equation is

${y_h} = {C_1}{e^{\sqrt 8 x}} + {C_2}{e^{ - \sqrt 8 x}} + {C_3}\cos \sqrt 8 x + {C_4}\sin \sqrt 8 x$

We will seek a particular solution in the form

$Y = (Ax + B)\cos x + (Cx + D)\sin x \Rightarrow Y' = A\cos x - (Ax + B)\sin x + C\sin x + (Cx + D)\cos x = (Cx + D + A)\cos x + ( - Ax - B + C)\sin x \Rightarrow$

$\Rightarrow Y'' = C\cos x - (Cx + D + A)\sin x - A\sin x + ( - Ax - B + C)\cos x = ( - Ax - B + 2C)\cos x + ( - Cx - D - 2A)\sin x \Rightarrow$

$\Rightarrow Y''' = - A\cos x - ( - Ax - B + 2C)\sin x - C\sin x + ( - Cx - D - 2A)\cos x = ( - Cx - D - 3A)\cos x + (Ax + B - 3C)\sin x \Rightarrow$

$\Rightarrow {Y^{(4)}} = - C\cos x - ( - Cx - D - 3A)\sin x + A\sin x + (Ax + B - 3C)\cos x = (Ax + B - 4C)\cos x + (Cx + D + 4A)\sin x$

Substitute into the equation:

$(Ax + B - 4C)\cos x + (Cx + D + 4A)\sin x - 64\left( {(Ax + B)\cos x + (Cx + D)\sin x} \right) = x\cos x$

$x\cos x(A - 64A) + \cos x(B - 4C - 64B) + x\sin x(C - 64C) + \sin x(D + 4A - 64D) = x\cos x$

$\left\{ \begin{array}{l} - 63A = 1\\ - 63B - 4C = 0\\ - 63C = 0\\ 4A - 63D = 0 \end{array} \right.$

$A = - \frac{1}{{63}},\,\,B = 0,\,C = 0,\,D = - \frac{4}{{3969}}$

Then

$Y = - \frac{1}{{63}}x\cos x - \frac{4}{{3969}}\sin x$

$y = {y_h} + Y = {C_1}{e^{\sqrt 8 x}} + {C_2}{e^{ - \sqrt 8 x}} + {C_3}\cos \sqrt 8 x + {C_4}\sin \sqrt 8 x - \frac{1}{{63}}x\cos x - \frac{4}{{3969}}\sin x$

Answer: $y = {C_1}{e^{\sqrt 8 x}} + {C_2}{e^{ - \sqrt 8 x}} + {C_3}\cos \sqrt 8 x + {C_4}\sin \sqrt 8 x - \frac{1}{{63}}x\cos x - \frac{4}{{3969}}\sin x$