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 = 1 cos x − 3x sin x

Answer 1

Given D.E is

$$y^{(4)} + 2y'' + y = \cos(x) − 3x\sin(x)$$

Thus, auxilary equation will be

$$m^4+2m^2+1=0\\ \implies (m^2+1)^2=0\\ \implies m=\pm i,\pm i$$

Hence,

$$C.F=(c_1+c_2x)\cos(x)+(c_3+c_4x)\sin(x)$$

Now, from $C.F$ we can guess the particular solution is

$$y_p(x)=(ax+bx^2)\cos(x)+(cx+dx^2)\sin(x)$$

Now, we have to find the value $a,b,c,d$ .

Thus, find the the expression

$$y_p^{(4)} + 2y_p'' + y_p$$

And compare the coefficients of the above expression with $\cos(x) − 3x\sin(x)$ we get

$$y_p^{(4)} + 2y_p'' + y_p=\cos(x)(-24ax-8b+24c)+\sin(x)(-24a-24cx-8d)\\\implies -24a=0\\ -8b+24c=1\\ -24c=3\\ -24a-8d=0$$

Thus,

$$a=d=0,b=\frac{1}{4},c=\frac{1}{8}\\$$

$$\implies y_p(x)=\frac{x^2}{4}\cos(x)+\frac{x^3}{8}\sin(x)$$