Find the general solution of y''' - 4y' = xe^(2x) + sin x + x

Find the general solution of $y''' - 4y' = xe^{2x} + \sin x + x$.

Step 1: Homogeneous solution

\[ r^3-4r=0 \implies r(r-2)(r+2)=0 \implies r_1=0,\ r_2=2,\ r_3=-2. \]

\[ Y_h = C_1 + C_2e^{2x} + C_3e^{-2x}. \]

Step 2: Split by superposition

Three unrelated forcing terms are present, so $y^*=y_1^*+y_2^*+y_3^*$.

Part 1: $f_1(x)=xe^{2x}$

Here $\alpha=2$, degree $n=1$. Since $r=2$ is a simple root ($s=1$), multiply the trial by $x$:

\[ y_1^* = x(Ax+B)e^{2x} = (Ax^2+Bx)e^{2x}. \]

Shift $D^3-4D$ by $D\to D+2$: $(D+2)^3-4(D+2) = D^3+6D^2+12D+8-4D-8 = D^3+6D^2+8D.$ So $Q=Ax^2+Bx$ must satisfy $Q'''+6Q''+8Q'=x$. With $Q'=2Ax+B,\ Q''=2A,\ Q'''=0$: $0+12A+8(2Ax+B)=x \implies 16Ax+(12A+8B)=x.$

\[ 16A=1\implies A=\tfrac1{16}, \qquad 12A+8B=0\implies \tfrac{12}{16}+8B=0\implies B=-\tfrac{3}{32}. \]

\[ y_1^* = \left(\frac{x^2}{16}-\frac{3x}{32}\right)e^{2x} = \frac{e^{2x}}{32}\left(2x^2-3x\right). \]

Part 2: $f_2(x)=\sin x$

Here $\alpha=0,\ \beta=1$; $r=\pm i$ is not a root (roots are $0,\pm2$), so $s=0$:

\[ y_2^*=A\cos x+B\sin x, \qquad y_2^{*\prime}=-A\sin x+B\cos x, \qquad y_2^{*\prime\prime}=-A\cos x-B\sin x, \qquad y_2^{*\prime\prime\prime}=A\sin x-B\cos x. \]

Substitute into $y_2^{*\prime\prime\prime}-4y_2^{*\prime}=\sin x$: \[(A\sin x-B\cos x)-4(-A\sin x+B\cos x)=\sin x \implies 5A\sin x-5B\cos x=\sin x.\]

\[ 5A=1\implies A=\tfrac15, \qquad -5B=0\implies B=0. \]

\[ y_2^* = \frac{\cos x}{5}. \]

Part 3: $f_3(x)=x$

Here $\alpha=0$, degree $n=1$. Since $r=0$ is a simple root ($s=1$), multiply the trial by $x$:

\[ y_3^* = x(Ax+B) = Ax^2+Bx, \qquad y_3^{*\prime}=2Ax+B, \qquad y_3^{*\prime\prime}=2A, \qquad y_3^{*\prime\prime\prime}=0. \]

Substitute into $y_3^{*\prime\prime\prime}-4y_3^{*\prime}=x$: $0-4(2Ax+B)=x \implies -8Ax-4B=x.$

\[ -8A=1\implies A=-\tfrac18, \qquad -4B=0\implies B=0. \]

\[ y_3^* = -\frac{x^2}{8}. \]

Step 3: General solution

Summing the three particular solutions with the homogeneous solution: