1
$\begingroup$

So I have $y' + 2y = \sin(\omega t)$.

Since $\sin(\omega t) = \operatorname{Im}(e^{i\omega t})$ I can solve the equation as far as to $$ y = \operatorname{Im}(\frac1{i\omega + 2}e^{i\omega t}) = \operatorname{Im}\left(\frac{2-i\omega}{\omega^2 + 2}\cdot(\cos(\omega t) + i\cdot \sin(\omega t))\right). $$

Now in my book they take it one step further, like this: $y = \frac 1{\omega^2 + 2}(2\sin(\omega t) - \omega \cos(\omega t))$.

That last step then becomes the final answer to the question, but I honestly cannot understand how they do that last jump. How do they remove the $\operatorname{Im}()$ thing and how does it affect the equation?

$\endgroup$
0
$\begingroup$

$$ \frac{2-i\omega}{\omega^2+2}(\cos \omega t + i \sin \omega t) = \frac{1}{\omega^2+2}(2\cos \omega t + 2i \sin \omega t -i\omega \cos\omega t + \omega \sin\omega t), $$ so the imaginary part is indeed $$ \frac{1}{\omega^2+2}(2 \sin \omega t -\omega \cos\omega t). $$

$\endgroup$
0
$\begingroup$

Just develop the product

$$\frac{2-i\omega}{\omega^2 + 2}\cdot (\cos{\omega t} + i\cdot \sin{\omega t})= {1\over \omega^2+2}\left[2\cos{\omega t}+\omega\sin{\omega t}+i\left(2\sin{\omega t}-\omega\cos{\omega t}\right)\right]$$

The imaginary part of $a+i\cdot b$ is $b$ and this gives the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.