x=e^t*sint y=e^t*cost 所以dx/dt=e^t*(sint +cost) ，dy/dt=e^t*(cost-sint)故dy/dx=(dy/dt) / (dx/dt)= (cost-sint) / (sint +cost)而d^2y/dx^2=d(dy/dx) /dt * dt/dx=d[(cost-sint) / (sint +cost)] /dt * dt/dx= [(-sint-cost)*(sint+cost) -(cost-sint)*(cost-sint)] /(sint+cost)^2 * 1/[e^t*(sint +cost)]= (-1-2sint*cost -1+2sint*cost)/[e^t*(sint+cost)^3]= -2 / [e^t*(sint+cost)^3]所以(d^2y/dx^2)*(x+y)^2= -2 / [e^t*(sint+cost)^3] * (e^t*sint +e^t*cost)^2= -2e^t /(sint+cost) 而2(x*dy/dx-y)=2[e^t*sint * (cost-sint) / (sint +cost) - e^t*cost]=2e^t *[sint * (cost-sint) -cost*(sint +cost)] /(sint +cost)=2e^t *[sint*cost -(sint)^2 -cost*sint -(cost)^2] / (sint +cost)= -2e^t /(sint+cost) 故(d^2y/dx^2)*(x+y)^2 =2(x*dy/dx-y) = -2e^t /(sint+cost)所以這兩個式子是相等的