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

Step-by-step solution

1. Write the homogeneous part: $y_h^{\prime \prime }+y_h=0$. Its characteristic equation $r^2+1=0$ gives $r=\pm i$.
2. Hence the complementary (homogeneous) solution is $y_h=C_1\cos x+C_2\sin x$.
3. For the particular solution, the RHS is $4\cos 2x$. Try $y_p=A\cos 2x+B\sin 2x$.
4. Compute $y_p^{\prime }=-2A\sin 2x+2B\cos 2x$ and $y_p^{\prime \prime }=-4A\cos 2x-4B\sin 2x$.
5. Substitute into the ODE: $(-4A\cos 2x-4B\sin 2x)+(A\cos 2x+B\sin 2x)=4\cos 2x$.
6. Combine like terms: $(-4A+A)\cos 2x+(-4B+B)\sin 2x=4\cos 2x$ $\Rightarrow$ $-3A\cos 2x-3B\sin 2x=4\cos 2x$.
7. Equate coefficients: $-3A=4\Rightarrow A=-\frac{4}{3}$, and $-3B=0\Rightarrow B=0$.
8. Thus $y_p=-\frac{4}{3}\cos 2x$.
9. The general solution is $y=y_h+y_p=C_1\cos x+C_2\sin x-\frac{4}{3}\cos 2x$.
10. Apply $y(0)=2$: $C_1\cdot 1+C_2\cdot 0-\frac{4}{3}\cdot 1=2\Rightarrow C_1=2+\frac{4}{3}=\frac{10}{3}$.
11. Differentiate the general solution: $y^{\prime }=-C_1\sin x+C_2\cos x+\frac{8}{3}\sin 2x$.
12. Apply $y^{\prime }(0)=-1$: $-C_1\cdot 0+C_2\cdot 1+0=-1\Rightarrow C_2=-1$.
13. Substitute $C_1$ and $C_2$ back: $y=\frac{10}{3}\cos x-\sin x-\frac{4}{3}\cos 2x$.
14. Combine constants: $\frac{10}{3}\cos x=2\cos x+\frac{4}{3}\cos x$. Adding the $-\frac{4}{3}\cos 2x$ term gives the simplified form $y=2\cos x-\sin x+\frac{4}{3}\cos 2x$.

Concept tested: Second‑order linear ODE with constant coefficients

Published 14 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked