Find the general solution of y''' + y'' = x^2 + 1 + 3e^(x)

Find the general solution of $y''' + y'' = x^2 + 1 + 3e^{x}$.

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: Split by superposition

The right-hand side is a sum of two unrelated forcing terms, so $y^*=y_1^*+y_2^*$, with each part treated separately.

Part 1: $f_1(x)=x^2+1$

Polynomial of degree $n=2$, $\alpha=0$. Since $r=0$ is a root of multiplicity $s=2$, the standard trial $Ax^2+Bx+C$ is multiplied by $x^2$: $y_1^* = x^2(Ax^2+Bx+C) = Ax^4+Bx^3+Cx^2.$

\[ y_1^{*\prime}=4Ax^3+3Bx^2+2Cx,\quad y_1^{*\prime\prime}=12Ax^2+6Bx+2C,\quad y_1^{*\prime\prime\prime}=24Ax+6B. \]

Substitute into $y_1^{*\prime\prime\prime}+y_1^{*\prime\prime}=x^2+1$: \[(24Ax+6B)+(12Ax^2+6Bx+2C)=x^2+1 \implies 12Ax^2+(24A+6B)x+(6B+2C)=x^2+1.\]

\[ x^2:\ 12A=1\implies A=\tfrac1{12}, \qquad x^1:\ 24A+6B=0\implies B=-\tfrac13, \qquad x^0:\ 6B+2C=1\implies C=\tfrac32. \]

\[ y_1^* = \frac{x^4}{12}-\frac{x^3}{3}+\frac{3x^2}{2}. \]

Part 2: $f_2(x)=3e^{x}$

Here $\alpha=1$, which is not a root of the characteristic equation (roots are $0,0,-1$), so $s=0$ and the trial is unmodified:

\[ y_2^*=Ae^{x} \implies y_2^{*\prime\prime\prime}+y_2^{*\prime\prime}=Ae^x+Ae^x=2Ae^x. \]

\[ 2Ae^x=3e^x \implies A=\tfrac32 \implies y_2^*=\frac{3e^{x}}{2}. \]

Step 3: Verify each part

Part 1: with $A=\tfrac1{12},B=-\tfrac13,C=\tfrac32$, $Q'''+Q''=(2x-2)+(x^2-2x+3)=x^2+1$. $\checkmark$
Part 2: $2\left(\tfrac32\right)e^x=3e^x$. $\checkmark$

Step 4: General solution

The solution is recorded in the result below.