Complex-Valued Solutions of Second-Order Linear Equations

[Dusty912]

Homework Statement

Suppose y(t) is a complex-valued solution of y'' +py' + qy=0 where p and q are real numbers. Show that if y(t)=y_re(t) + iy_im(t), where y_re(t) and y_im(t) are real valued, then both y_re(t) and y_im(t) are solutions of the second-order equation.

Homework Equations

We can use the idea that a complex numbers is zero as long as it's real and imaginary parts are equal to zero.

The Attempt at a Solution

I do not have an attempt at this solution. I do not know where to begin.

thanks for your help :)

 

[LCKurtz]

Homework Statement

Suppose y(t) is a complex-valued solution of y'' +py' + qy=0 where p and q are real numbers. Show that if y(t)=y_re(t) + iy_im(t), where y_re(t) and y_im(t) are real valued, then both y_re(t) and y_im(t) are solutions of the second-order equation.

Homework Equations

We can use the idea that a complex numbers is zero as long as it's real and imaginary parts are equal to zero.

The Attempt at a Solution

I do not have an attempt at this solution. I do not know where to begin.

thanks for your help :)

Begin by plugging y(t)=y_re(t) + iy_im(t) into your DE and see what happens.

 

[Dusty912]

then I would get y''_re(t) + py'_re(t) +qy_re(t) +y''_im(t) +py'_im(t) +qy_im(t)=0

so the real part would add up too zero and the imaginary part would add up to zero.

 

[Dusty912]

would this be correct?

 

[vela]

Where'd the factor of $i$ go?

 

[Dusty912]

oops so it would be y''_re(t) + py'_re(t) +qy_re(t) +iy''_im(t) +ipy'_im(t) +iqy_im(t)=0

 

[Hiranya Pasan]

lets take yre as a solution, y''re + py're+qyre=0
then take yim as a solution, y''im+py'im +qyim=0
the multiply with i both sides of the 2nd eqn,
i(y''im+py'im +qyim)=0

linear combination of particular solutions isalso a solution,

general solution y(t)=yre(t) + i yim(t)