Find the general solution of y''' - y'' = xe^(x)

Find the general solution of $y''' - y'' = xe^{x}$.

Step 1: Homogeneous solution

\[ r^3-r^2=0 \implies r^2(r-1)=0 \implies r_1=r_2=0\ \text{(double)},\quad r_3=1. \]

\[ Y_h = C_1 + C_2x + C_3e^{x}. \]

Step 2: Choose the trial form

The right-hand side is $xe^{\alpha x}$ with $\alpha=1$, a polynomial of degree $n=1$ times $e^x$. Since $r=1$ is a simple root of the characteristic equation (multiplicity $s=1$), the standard trial $(Ax+B)e^x$ must be multiplied by $x^s=x$:

\[ y^* = x(Ax+B)e^{x} = (Ax^2+Bx)e^{x}. \]

Step 3: Exponential-shift substitution

Write $y^*=Qe^x$ with $Q=Ax^2+Bx$. Shifting the operator $D^3-D^2$ by $D\to D+1$: $(D+1)^3-(D+1)^2 = D^3+3D^2+3D+1-D^2-2D-1 = D^3+2D^2+D.$ So $Q$ must satisfy $Q'''+2Q''+Q'=x$. With $Q=Ax^2+Bx$: \[Q'=2Ax+B, \qquad Q''=2A, \qquad Q'''=0.\]

Substitute: $0+2(2A)+(2Ax+B)=x \implies 2Ax+(4A+B)=x.$

Step 4: Match coefficients

\[ x^1:\ 2A=1\implies A=\tfrac12, \qquad x^0:\ 4A+B=0\implies 2+B=0\implies B=-2. \]

\[ y^* = \left(\frac{x^2}{2}-2x\right)e^{x} = \frac{e^{x}}{2}\left(x^2-4x\right). \]

Step 5: Verify

With $A=\tfrac12,\ B=-2$: $Q'''+2Q''+Q'=0+2(1)+(x-2)=x$, confirming the trial satisfies the shifted equation. $\checkmark$

Step 6: General solution

The solution is recorded in the result below.