Question

Find a particular solution of the given differential equation. Use a CAS as an aid in carrying out differentiations, simplifications, and algebra.
y(4) + 2y'' + y = 6 cos x − 8x sin x

yp =

Answer 1

$\lambda^4+2\cdot \lambda^2+1=0\\ (\lambda^2+1)^2=(\lambda+i)^2\cdot (\lambda-i)^2=0\\ \lambda_1=\lambda_2=-i, \lambda_3=\lambda_4=i\\ y_{h}=(c_1+c_2x)sin(x)+(c_3+c_4x)cos(x)\\ y_0=(\frac{1}{4}x^2-\frac{9}{16})cos(x)+(\frac{1}{3}x^3-\frac{1}{2}x)sin(x).\\ y(x)=y_{h}+y_{0}=(c_1+c_2x)sin(x)+(c_3+c_4x)cos(x)+ (\frac{1}{4}x^2-\frac{9}{16})cos(x)+(\frac{1}{3}x^3-\frac{1}{2}x)sin(x);$