Solve the second‑order linear differential equation $y^{\prime \prime }+4y=\sin 2x$ with the initial conditions $y(0)=0$ and $y^{\prime }(0)=1$.

Step-by-step solution

1. Write the associated homogeneous equation: $y^{\prime \prime }+4y=0$.
2. Solve the characteristic equation $r^2+4=0$ giving roots $r=\pm 2i$.
3. Hence the complementary (homogeneous) solution is $y_c=C_1\cos 2x+C_2\sin 2x$.
4. For the particular solution, note that the forcing term $\sin 2x$ is a solution of the homogeneous part, so try $y_p=x(A\cos 2x+B\sin 2x)$.
5. Compute $y_p^{\prime }$ and $y_p^{\prime \prime }$ and substitute into $y^{\prime \prime }+4y=\sin 2x$.
6. After simplification, the coefficients give $A=0$ and $B=\frac{1}{2}$. Thus $y_p=\frac{x}{2}\sin 2x$.
7. Combine: $y(x)=C_1\cos 2x+C_2\sin 2x+\frac{x}{2}\sin 2x$.
8. Apply the initial condition $y(0)=0$: $C_1=0$ because $\cos 0=1$ and the other terms vanish at $x=0$.
9. Differentiate $y$ to get $y^{\prime }(x)=-2C_1\sin 2x+2C_2\cos 2x+\frac{1}{2}\sin 2x+x\cos 2x$.
10. Use $y^{\prime }(0)=1$: $2C_2=1$ ⇒ $C_2=\frac{1}{2}$.
11. Substitute $C_1=0$ and $C_2=\frac{1}{2}$ back: $y(x)=\frac{1}{2}\cos 2x+\frac{x}{2}\sin 2x$.
12. Combine the terms using the identity $\cos 2x=\frac{d}{dx}(\sin 2x)/2$ or recognise that the expression simplifies to $y(x)=\frac{x}{2}\cos 2x$.
13. Thus the final solution satisfying the initial conditions is $y(x)=\frac{x}{2}\cos 2x$.

Concept tested: Second‑order linear non‑homogeneous ODE

Published 8 September 2026 · How this solution was prepared and checked