Find the general solution of y''' + y'' - y' + 15y = sin 2x

Find the general solution of $y''' + y'' - y' + 15y = \sin 2x$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$ into the homogeneous equation gives the cubic

\[ r^3+r^2-r+15=0. \]

Step 2: Finding a rational root

By the Rational Root Theorem, candidates divide $15$. Testing $r=-3$: $(-3)^3+(-3)^2-(-3)+15 = -27+9+3+15 = 0.$ $\checkmark$ So $r=-3$ is a root, and $r^3+r^2-r+15=(r+3)(r^2-2r+5)$; expanding confirms $(r+3)(r^2-2r+5)=r^3-2r^2+5r+3r^2-6r+15=r^3+r^2-r+15$.$\ \checkmark$

Step 3: Remaining roots

\[ r^2-2r+5=0 \implies r=\frac{2\pm\sqrt{4-20}}{2}=1\pm2i. \] So the three characteristic roots are $r_1=-3$, $r_{2,3}=1\pm2i$.

Step 4: Homogeneous solution

\[ Y_h = C_1e^{-3x} + e^{x}\bigl(C_2\cos2x + C_3\sin2x\bigr). \]

Step 5: Resonance check and trial form

The forcing term $\sin2x$ corresponds to $\alpha=0$, $\beta=2$, i.e.\ complex exponent $2i$. Comparing against the roots $\{-3,\,1\pm2i\}$: $2i$ is not among them ($1+2i\neq2i$), so there is no resonance ($s=0$) and the trial is the plain

\[ y^{*} = A\cos2x + B\sin2x. \]

Step 6: Derivatives

\[ y^{*\prime}=-2A\sin2x+2B\cos2x,\qquad y^{*\prime\prime}=-4A\cos2x-4B\sin2x,\qquad y^{*\prime\prime\prime}=8A\sin2x-8B\cos2x. \]

Step 7: Substitute and match coefficients

Substituting into $y'''+y''-y'+15y=\sin2x$ and collecting the $\cos2x$ and $\sin2x$ coefficients: $\cos2x:\ -8B-4A-2B+15A = 11A-10B, \qquad \sin2x:\ 8A-4B+2A+15B = 10A+11B.$

Matching against the right side $0\cdot\cos2x+1\cdot\sin2x$: \[11A-10B=0, \qquad 10A+11B=1.\]

Step 8: Solve the linear system

From the first equation, $A=\dfrac{10B}{11}$. Substituting into the second: \[10\cdot\frac{10B}{11}+11B=1 \implies \frac{100B}{11}+\frac{121B}{11}=1 \implies \frac{221B}{11}=1 \implies B=\frac{11}{221}.\]

\[ A = \frac{10}{11}\cdot\frac{11}{221} = \frac{10}{221}. \]

Step 9: Particular solution

\[ y^{*} = \frac{10\cos2x+11\sin2x}{221}. \]

Step 10: General solution

The solution is recorded in the result below.