Solving for Imaginary Solutions in Second-Order Differential Equations

[cragar]

Homework Statement

Show that the imaginary part of the solution of
$z''+z'+z=te^{it}$ is a solution of $y''+y'+y=tsin{t}$

The Attempt at a Solution

Ok so I first make the guess that $z(t)=(at+b)e^{it}$
then I find z' and z'' and plug it back in and then equate the coefficients of t and then all the leftover constants.

I do this and I get a=-i and b=(2i+1)
so then I plug this in back to the original guess for z(t) and then multiply it by Eulers formula
and then take the imaginary part and see if it works for y(t). Is this the right approach.
I seem to be off by a cosine factor, I could post my work, but I just wanted to know if this is the right approach.

 

[Curious3141]

Homework Statement

Show that the imaginary part of the solution of
$z''+z'+z=te^{it}$ is a solution of $y''+y'+y=tsin{t}$

The Attempt at a Solution

Ok so I first make the guess that $z(t)=(at+b)e^{it}$
then I find z' and z'' and plug it back in and then equate the coefficients of t and then all the leftover constants.

I do this and I get a=-i and b=(2i+1)
so then I plug this in back to the original guess for z(t) and then multiply it by Eulers formula
and then take the imaginary part and see if it works for y(t). Is this the right approach.
I seem to be off by a cosine factor, I could post my work, but I just wanted to know if this is the right approach.

Frankly, I think you're working too hard.

Why don't you just try putting $z(t) = f(t) + ig(t)$ into the LHS of the original equation, where f(t) and g(t) are both real-valued functions of t? Then express the RHS as $t\cos t + it\sin t$. You will immediately see, by equating the imaginary parts, that g(t) has to be a solution of the second equation.

 

[cragar]

ya that way is more slick, thanks for the help