Find the general solution of y''' + y' = x^4

Find the general solution of $y''' + y' = x^4$.

Step 1: Homogeneous Solution

\[ r^3+r=0 \implies r(r^2+1)=0 \implies r_1=0,\ r_{2,3}=\pm i. \]

\[ Y_h = C_1 + C_2\cos x + C_3\sin x. \]

Step 2: Particular Solution

The right-hand side $f(x)=x^4$ is a polynomial of degree 4. Here $r=0$ is a root of the characteristic equation — a simple one, of multiplicity $s=1$ — so the standard degree-4 trial polynomial must be multiplied by $x^s=x$ to avoid duplicating a term already present in $Y_h$:

\[ y^* = x\left(Ax^4+Bx^3+Cx^2+Dx+E\right) = Ax^5+Bx^4+Cx^3+Dx^2+Ex. \]

\[ y^{*\prime}=5Ax^4+4Bx^3+3Cx^2+2Dx+E, \quad y^{*\prime\prime}=20Ax^3+12Bx^2+6Cx+2D, \quad y^{*\prime\prime\prime}=60Ax^2+24Bx+6C. \]

Substitute into $y'''+y'=x^4$: $(60Ax^2+24Bx+6C)+(5Ax^4+4Bx^3+3Cx^2+2Dx+E)=x^4$, $5Ax^4+4Bx^3+(60A+3C)x^2+(24B+2D)x+(6C+E)=x^4$.

Step 3: Match Coefficients

\[ \begin{aligned} x^4:&\ 5A=1 \implies A=\tfrac{1}{5}, \\ x^3:&\ 4B=0 \implies B=0, \\ x^2:&\ 60A+3C=0 \implies 12+3C=0 \implies C=-4, \\ x^1:&\ 24B+2D=0 \implies D=0, \\ x^0:&\ 6C+E=0 \implies -24+E=0 \implies E=24. \end{aligned} \]

\[ y^* = \frac{x^5}{5} - 4x^3 + 24x. \]

Step 4: Verification

\[ y^{*\prime}=x^4-12x^2+24, \qquad y^{*\prime\prime}=4x^3-24x, \qquad y^{*\prime\prime\prime}=12x^2-24. \]

\[ y^{*\prime\prime\prime}+y^{*\prime} = (12x^2-24)+(x^4-12x^2+24) = x^4.\ \checkmark \]

Step 5: General Solution