Using method of Undetermined coefficients to solve $y'''' + y'' = 3x^2 + 4\sin x - 2\cos x$.

Question

Here is the differential eqn I am having trouble with

$$y'''' + y'' = 3x^2 + 4\sin x - 2\cos x$$

So I am supposed to solve this by method of undetermined coefficients.

So the characteristic equation should be $m^4 + m^2 = 0$

so $m=0$ (repeated) and $m=\pm i$ (euler identity)

$y_h = c_1 + (A\cos x + B\sin x)$

Now to find the complete solution I need to find yp.

So according to the Sum rule $y =y_h + y_{p1} + y_{p2}$

to get $y_{p1}$,

So I am now supposed to first solve $y'''' + y'' = 3x^2$

So let a sol be

$$y_{p1} = K_2x^2 + K_1x + K_0$$ So, \begin{align*} y_{p1}'&= 2K_2x + K_1\\ y_{p1}''&= 2K_2\\ y_{p1}''''&= 0 \end{align*} This is the part where I get stuck Substituting values of y to find the coefficients we get $$2K_2 = 3x^2$$

$$K_2 = 0?$$ There must be something wrong here.

NOTE: I am following the process of solving higher order ODEs as mentioned in the book Advanced Engineering Mathematics by Erwin Kreyszig 9th edition.

Sorry for the bad formatting, this is the first time I am posting here.

Answer 1

First since $m=0$ is of multiplicity two the homogeneous solution is $$y_h=c_1+c_2x+c_3 \cos x+c_4 \sin x$$ Now for the particular solution $$y'''' + y'' = 3x^2$$ Susbtitute $y''=z$ then the equation becomes $$z'' + z= 3x^2$$ Use $z_p=Ax^2+Bx+C$ as a particular solution Then you find $$z_p=3x^2-6$$ Integrate twice to find $y_p$ $$y_p''=3x^2-6$$

Edit second part I used $$z_p=x(A\cos x + B \sin x)$$ For this equation $$z''+z=4 \sin x -2 \cos x$$ Then I got $$(A,B)=(-2,-1)$$ Integrate by part but take first part of the integration since the second part is absorbed by the homogeneous solution $$y''_p=-x \sin x -2x \cos x$$ First integration $$y'_p=x \cos x -2 x\sin x$$ Second integration. Finally: $$\boxed {y_p=x \sin x +2 x\cos x}$$

Answer 2

Since $m=0$ is a root of the characteristic equation and their multiplicity is $2$, and $\pm i$ are two complex conjugated roots it follows $$y_h=c_1+c_2x+c_3\cos x+c_4\sin x$$ At a first glance we can think about $y_{p1}=Ax^2+Bx+C$ and $y_{p2}=E\cos x+F\sin x$, but these two particular solutions contains solutions of the associated homogeneous DE, so we must to modify them properly: $$y_{p1}=\color{red}{x^2}\left(Ax^2+Bx+C\right)\qquad\text{and}\qquad y_{p2}=\color{red}{x}\left(E\cos x+F\sin x\right)$$