Question

Solution of (D^3 -D)y = x^3 is

Answer 1

Since the indicated equation is inhomogeneous, the solution consists of two parts

$$y(x)=y_{hom}(x)+y_{inhom}(x),\,\,\,\text{where}\\[0.3cm] \left[\begin{array}{l} \left(D^3-D\right)y=0\longrightarrow\,\,\,y=y_{hom}(x)\\[0.3cm] \left(D^3-D\right)y=x^3\longrightarrow\,\,\,y=y_{inhom}(x),\,\,\,\text{some solution} \end{array}\right.$$

1 STEP: solve a homogeneous equation

$$\left(D^3-D\right)y=0\\[0.3cm] \text{We are looking for a solution in the form}\,\,\, y(x)=e^{kx}\\[0.3cm] Dy=\frac{d(e^{kx})}{dx}=k\cdot e^{kx}\\[0.3cm] D^3y=\frac{d^3(e^{kx})}{dx^3}=k^3\cdot e^{kx}\\[0.3cm] \left(D^3-D\right)y=0\longrightarrow (k^3-k)\cdot e^{kx}=0\\[0.3cm] k\cdot(k-1)\cdot(k+1)=0\longrightarrow\left[\begin{array}{l} k=0\\[0.3cm] k=1\\[0.3cm] k=-1 \end{array}\right.$$

Conclusion,

$$\boxed{y_{hom}(x)=C_1\cdot e^{x}+C_2+C_3\cdot e^{-x}}$$

2 STEP: we are looking for a particular solution to the inhomogeneous equation

$$\left(D^3-D\right)y=x^3$$

We are looking for a solution in the form

$$y_{inhom}(x)=Ax^4+Bx^3+Cx^2+Dx+E\longrightarrow\\[0.3cm] \left[\begin{array}{l} Dy=\displaystyle\frac{dy_{inhom}}{dx}=4Ax^3+3Bx^2+2Cx+D\\[0.3cm] D^3y=\displaystyle\frac{d^3y_{inhom}}{dx^3}=24Ax+6B \end{array}\right.$$

Then,

$$\left(D^3-D\right)y=x^3\longrightarrow\\[0.3cm] (24Ax+6B)-(4Ax^3+3Bx^2+2Cx+D)=x^3\longrightarrow\\[0.3cm] (-1-4A)x^3-3Bx^2+(24A-2C)x+(6B-D)=0\longrightarrow\\[0.3cm] \left[\begin{array}{l} x^3\,\,:\,\,-1-4A=0\longrightarrow\boxed{A=-\displaystyle\frac{1}{4}}\\[0.3cm] x^2\,\,:\,\,-3B=0\longrightarrow\boxed{B=0}\\[0.3cm] x^1\,\,:\,\,24A-2C=0\longrightarrow\boxed{C=-3}\\[0.3cm] x^0\,\,:\,\,6B-D=0\longrightarrow\boxed{D=0}\\[0.3cm] \end{array}\right.$$

Conclusion,

$$\boxed{y_{inhom}(x)=-\frac{x^4}{4}-3x^2}$$

General conclusion,

$$y(x)=y_{hom}(x)+y_{inhom}(x)\longrightarrow\\[0.3cm] \boxed{y(x)=C_1\cdot e^{x}+C_2+C_3\cdot e^{-x}-\frac{x^4}{4}-3x^2}$$

ANSWER

$$y(x)=C_1\cdot e^{x}+C_2+C_3\cdot e^{-x}-\frac{x^4}{4}-3x^2$$