Solution. The characteristic equation for a ordinary differential equation
k 2 + k + 1 = 0 k^2+k+1=0 k 2 + k + 1 = 0 Solve the quadratic equation
D = 1 2 − 4 × 1 × 1 = − 3 < 0 D=1^2-4\times 1 \times1=-3<0 D = 1 2 − 4 × 1 × 1 = − 3 < 0 Roots of the equation
k 1 = − 1 − 3 i 2 k_1=\frac{-1-\sqrt{3}i}{2} k 1 = 2 − 1 − 3 i
k 2 = − 1 + 3 i 2 k_2=\frac{-1+\sqrt{3}i}{2} k 2 = 2 − 1 + 3 i Therefore the solution for a ordinary differential equation
y = e − x 2 ( C 1 s i n ( 3 x 2 ) + C 2 c o s ( 3 x 2 ) ) y=e^{\frac{-x}{2}}(C_1sin(\frac{\sqrt{3}x}{2})+C_2cos(\frac{\sqrt{3}x}{2})) y = e 2 − x ( C 1 s in ( 2 3 x ) + C 2 cos ( 2 3 x )) where C1 and C2 are constants.
Answer.
y = e − x 2 ( C 1 s i n ( 3 x 2 ) + C 2 c o s ( 3 x 2 ) ) y=e^{\frac{-x}{2}}(C_1sin(\frac{\sqrt{3}x}{2})+C_2cos(\frac{\sqrt{3}x}{2})) y = e 2 − x ( C 1 s in ( 2 3 x ) + C 2 cos ( 2 3 x )) where C1 and C2 are constants.
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