Differential Equations old and the new

[jlatshaw]

Homework Statement

Given:
y''[t] + 25 y[t] = 0

I know that the solution to this DE is of the form:

y[t] = K1 E^(-5 i t) + K2 ​​E ^(5 i t)

I get that, that makes sense to me, however when I look in old DE books I see the solution to the same problem written as:


C1 Cos[5 t] + C2 Sin[k t]


How do they get this, and how can I establish a transition between the two?
How are they related?

Thanks!
-James

 

[Ray Vickson]

Homework Statement

Given:
y''[t] + 25 y[t] = 0

I know that the solution to this DE is of the form:

y[t] = K1 E^(-5 i t) + K2 ​​E ^(5 i t)

I get that, that makes sense to me, however when I look in old DE books I see the solution to the same problem written as:


C1 Cos[5 t] + C2 Sin[k t]


How do they get this, and how can I establish a transition between the two?
How are they related?

Thanks!
-James

Surely you must have seen $e^{ix} = \cos x + i \sin x.$ It is one of the most fundamental results you will ever meet.

 

[jlatshaw]

Ok, I guess it's my constants that are getting me in a twist.

Where am I going wrong here?

knowing that:
E^(ix) = cos(x) + i sin(x)
and
E^(-ix) = cos(x) - i sin(x)

K1 E^(-5 i t) + K2 ​​E ^(5 i t)
(Let x = 5t)
= K1 (cos(x) - i sin(x)) + K2(cos(x) + i sin(x))
= cos(x) (K1 + K2) + i sin(x) (K2-k1)

So what should I do from here?
Or am I just really over thinking this one?

 

[Ray Vickson]

Ok, I guess it's my constants that are getting me in a twist.

Where am I going wrong here?

knowing that:
E^(ix) = cos(x) + i sin(x)
and
E^(-ix) = cos(x) - i sin(x)

K1 E^(-5 i t) + K2 ​​E ^(5 i t)
(Let x = 5t)
= K1 (cos(x) - i sin(x)) + K2(cos(x) + i sin(x))
= cos(x) (K1 + K2) + i sin(x) (K2-k1)

So what should I do from here?
Or am I just really over thinking this one?

So you have $K_1+K_2 = C_1$ and $i (K_2 - K_1) = C_2.$ You are over-thinking it: a linear combinations of sin(x) and cos(x) can be written as a linear combination of exp(ix) and exp(-ix). Either way of writing it is fine. And, BTW: there is nothing "new" about using complex exps and nothing "old" about using sin and cos: you just fit the form to the problem you are addressing. Some problems need the sin and cos form; others cry out to be left in exp form.

 

[jlatshaw]

oh ok.
One final question regarding this.
When I switch up the constants, I get
C1 Cos(5t) + C2 Sin(5t)
But, I'm suppose to come up with
C1 Cos[5 t] + C2 Sin[k t]

So what should I do about that k? Why is it not a 5? Or does k have to equal 5?

Thanks.
-James

 

[Ray Vickson]

oh ok.
One final question regarding this.
When I switch up the constants, I get
C1 Cos(5t) + C2 Sin(5t)
But, I'm suppose to come up with
C1 Cos[5 t] + C2 Sin[k t]

So what should I do about that k? Why is it not a 5? Or does k have to equal 5?

Thanks.
-James

You tell me.

 

[jlatshaw]

Well, I want to say that:
C1 Cos(5t) + C2 Sin(5t) = C1 Cos(5t) + C2 Sin(kt)
and that k is just 5.
Is that right to say?

I'm asking this not because I have trouble determining if k = 5, but mainly to make sure I did my work correct in finding
C1 Cos(5t) + C2 Sin(5t)
then I know that k = 5.
But then that just makes me confused because how would of they solved for it by having a k there and not knowing that k = 5.

 

[Ray Vickson]

Well, I want to say that:
C1 Cos(5t) + C2 Sin(5t) = C1 Cos(5t) + C2 Sin(kt)
and that k is just 5.
Is that right to say?

I'm asking this not because I have trouble determining if k = 5, but mainly to make sure I did my work correct in finding
C1 Cos(5t) + C2 Sin(5t)
then I know that k = 5.
But then that just makes me confused because how would of they solved for it by having a k there and not knowing that k = 5.

I don't get it: you know that sin(5t) and cos(5t) are solutions, so k must be 5. End of story.

If you want to make a lot of unnecessary work for yourself, you can plug in sin(kt) into the DE and see if it works; it will work if you choose k properly.