# Question:2nd order ODE with complex coefficient

## Question:2nd order ODE with complex coefficient

Maple

Hi

Assume a linear second order ode with constant coefficients as follows:

M*u''+C*u'+K*u=0

where the symbol (') denotes derivative with respect to time and M, C and K are positive real constants.

The initial conditions are u(0)=u0 and u'(0)=u'0.

Substituting u=exp(a*t) in the ODE to calculate characteristic equation, one has

M*a^2+C*a+K=0  ---> If 0<C<2sqrt(M*K) then u=exp(-C/(2M)*t)*(c1*sin(w*t)+c2*cos(W*t))

in which c1 and c2 can be obtained from initial conditions and W=sqrt(C^2-4*M*K)/2M.

For the case that C is imaginary number, assuming C=i*c yields

{a1,a2} ={ -(c/(2M) +sqrt(c^2+4*M*K)/2M)*i,(-c/(2M) +sqrt(c^2+4*M*K)/2M)*i }

where i is one of the square roots of -1.

Is it true to write u=c1*(sin(a1*t)+cos(a1*t))+c2*(sin(a2*t)+cos(a2*t)) ?

I solve two examples by Maple,

dsolve({diff(u(t), t, t)+0.1*(diff(u(t), t))+2*u(t)} union {u(0) = 1, (D(u))(0) =0.1});

dsolve({diff(u(t), t, t)+0.1*I*(diff(u(t), t))+2*u(t)} union {u(0) = 1, (D(u))(0) = 0.1});

Second example gives complex answer. Is it possible to get trigonometric answer with real constants c1 and c2?

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