Find a complex number from differential equations.

[tomeatworld]

Homework Statement

Find complex number \lambda such that e^\lambdat solves
\frac{d^{2}y}{dt^{2}} + 4\frac{dy}{dt} + 5y = 0

Express this solution in the form e^at(cos(bt) + i sin(bt))

Homework Equations

The Attempt at a Solution

So the first part is fine, using \lambda² + 4\lambda + 5 = 0 to get values of \lambda at -2\pmi. From here, I've been taught to use:

y = Ae^\lambda₁t + Be^\lambda₂t but this time, it doesn't help get to the required form.

Using what I've mention, I can get to A=-2 but finding B seems to be a mystery. Any help greatly appreciated!

 

[l'Hôpital]

Are you aware of Euler's identity?
<br /> e^{ix} = \cos x + i \sin x<br />

 

[tomeatworld]

I am. Converting to (and from) polar form. My attempt went as far as:

y = Ce^{-2t+it} + De^{-2t-it}
y = e^{-2t}(Ce^{it}+De^{-it})

That's how I got to A=-2 but from here:

y =e^{-2t} (C(cos(it) + i sin(it)) + D(cos(-it) + i sin(it)))

and from there, getting a value for the B in question is eluding me.
Note: I've changed the constants in the equation so they don't clash with the required values.

 

[vela]

I think the problem is just asking you to express each of the solutions in the form given separately, before you get to the step of writing down with the general solution.

 

[tomeatworld]

it was just the question was phrased:
"Find complex number \lambda such that e^\lambdat is a solution to the equation (see above). Express this solution in the form (see above) and give the values of real numbers A and B"
Just that seemed like there was one :/

 

[l'Hôpital]

<br /> y = e^{-2t} (C(\cos(it) + i \sin(it)) + D(\cos(-it) + i \sin(-it)))<br />

Note the following:

<br /> \cos -x = \cos x<br />
<br /> \sin -x = - sin x<br />

So...

<br /> y = e^{-2t}[ (C+D) \cos (it) + (C -D) \sin(it)]<br />

We can let C+D and C-D be constant by themselves, say K and M
<br /> y = e^{-2t}( K \cos (it) + M \sin(it))<br />
This is all just to make it pretty and for future problems. For this particular problem, all that you needed to know was that sine was odd and cosine even.
Thus, the b value is, of course, one.

 

[tomeatworld]

i see! right, so maybe a quick sketch of sine and cosine graphs would have cleared this up for me! I'll remember that in future! Thanks a load for all the help!

 

[HallsofIvy]

I am. Converting to (and from) polar form. My attempt went as far as:

y = Ce^{-2t+it} + De^{-2t-it}
y = e^{-2t}(Ce^{it}+De^{-it})

That's how I got to A=-2 but from here:

y =e^{-2t} (C(cos(it) + i sin(it)) + D(cos(-it) + i sin(it)))

No! e^{it}= cos(t)+ i sin(t), not e^{it}= cos(it)+ i sin(it)

y= e^{-2t}(C e^{it}+ De^{-it})= e^{-2t}(Ccos(t)+ Ci sin(t)+ D cos(-t)+ Di sin(-t))
Now, cos(-t)= cos(t) and sin(-t)= -sin(t) so this is
y= e^{-2t}(C cos(t)+ Ci sin(t)+ D cos(t)- Di sin(t))= e^{-2t}((C+D)cos(t)+ (C-D)i sin(t))

and from there, getting a value for the B in question is eluding me.



Note: I've changed the constants in the equation so they don't clash with the required values.

 

[tomeatworld]

Of course. Sorry, slip of the keyboard ;)