Question 3.3.4: Solve d^4y/dx^4 + 2d²y/dx² + y = 0....

The auxiliary equation m^4+2 m^2+1=\left(m^2+1\right)^2=0 has roots m_1=m_3=i and m_2=m_4=-i \text {. } Thus from Case II the solution is

y=C_1 e^{i x}+C_2 e^{-i x}+C_3 x e^{i x}+C_4 x e^{-i x} .

By Euler’s formula the grouping C_1 e^{i x}+C_2 e^{-i x} can be rewritten as c_1 \cos x+c_2 \sin x after a relabeling of constants. Similarly x\left(C_3 e^{i x}+C_4 e^{-i x}\right) can be expressed as x\left(c_3 \cos x+c_4 \sin x\right).

Hence the general solution is

y=c_1 \cos x+c_2 \sin x+c_3 x \cos x+c_4 x \sin x .