= Solution
The <characteristic equation> has roots $-3\pm4i$, giving the <homogeneous solution> $e^{-3x}(A\cos4x+B\sin4x)$. For the sinusoidal <forcing term>, $y_p=\sin5x$ is a <particular solution>, since its second <derivative> cancels $25y_p$ and $6y_p'=30\cos5x$. The zero initial displacement gives $A=0$; zero initial <derivative> gives $4B+5=0$. Thus
$$
\boxed{y(x)=\sin5x-\frac54e^{-3x}\sin4x.}
$$
The <transient response> decays as $e^{-3x}$, leaving the <steady-state response> $\sin5x$. Near zero, $y(0)=y'(0)=0$ and $y''(0)=30$, so the sketch starts with $y\sim15x^2$ and a horizontal <tangent line>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2-driven-oscillation.png]
{title=The decaying transient brings the initially stationary solution to a sinusoidal response}
{description=Solid: the complete solution. Dashed: the surviving $\sin5x$ term. The lower panel resolves the exponentially decaying difference.}
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