题图引用的是二阶常系数齐次线性微分方程求通解的基本公式。特征方程的2个根 r1、r2为共轭复根 r1、2 =a ± bi 时,通解y = [ e^(ax) ][ C1cos(bx) + C2sin(bx) ];本题便是y = [ e^(-x/2) ][C1cos(√3x/2) + C2sin(√3x/2) ]通解公式推导详见高数。简叙如下:y = C1y1 + C2y2 = C1{e^[ (a+bi)x ] } + C2{ e^[ (a-bi)x ] };为得到实值通解,用欧拉公式;y1 = e^[ (a+bi)x ] = [ e^(ax) ][e^(bix) ] =[ e^(ax) ][ cos(bx) + isin(bx) ]y2 = e^[ (a-bi)x ] = [ e^(ax) ][ e^(-bix) ] = [ e^(ax) ][ cos(bx) - isin(bx) ]y1' = ( y1 + y2 )/2 =[ e^(ax) ]cos(bx);y2' =( y1 - y2 )/(2i) = [ e^(ax) ]sin(bx);y1'、y2'也是微分方程的解,且yi'/y2' = cot(bx),不是常数;故通解y = C1y1' + C2y2' =[ e^(ax) ][ C1cos(bx) + C2sin(bx) ] 。