Jaller's homework

I hope my deduction is correct.If anybody wants to learn more about trigonometric functions go here:

http://en.wikipedia.org/wiki/Trigonometric_functions

ouchh

I wish I was working on Trigonometric functions again.  Right now, I am having lots of fun with them in calculus.  There is nothing like taking the fourth derivative of a four term trigonometric function.  I have never used so much paper for one class.

Can you give me the problem pls? I would like to try it...

Let me see if I can find it.

Okay here it is.  It is relatively simple, but it takes quite a bit of space.  Sorry for the representation, but this is the best I could do.

Determine whether the function is a solution of the differential equation 4th derivative of y - 16y = 0

y = C(subscript 1)e^(2x) + C(subscript 2)e^(-2x) + C(subscript 3)sin(2x) + C(subscript 4)cos(2x)

I think that solving the differential equation should be simpler.I tried with Euler's method (it was really helpful that y is given)

The characteristic equation is: r^4-16r=0 => r(r^3-16)=0

 r1=0 ; the solutions of r^3-16=0 are real numbers (r^3=16 is a bynomial equation (I dunno the exact name...) so the solutions are r2=r3=r4 = 16^(1/3))

Let's go the other way:

You can deduce the solutions for the characteristic equation from the form of y if it is given:

y=C1y1+C2y2+C3y3+C4y4

y1=e^(r1x) =>r1=2   

y2=e^(r2x) =>r2=-2

for the rest let's give the general form of a complex number: Z=A+Bi

If a complex number Z is solution of an equation then the conjugate of Z is also solution

y3=e^(Ax)sinBx =>A=0;B=2 =>Z=2i

y4=e^(Ax)cosBx=>the conjugate of Z=-2i which is true because the cosine function is an even function and cos(-x) = cos(x)

So the characteristic equation r^4 - 16r = 0 should admit the solutions r1=2, r2=-2; r3=2i; r4=-2i which, I think, it is not true.

can anybody confirm if I am right...or wrong??

I can post a solution, although I did not use Euler's method.  Give me a few minutes to work through the problem.

ok..thank you

For reading purposes, the fourth derivative will be represented by y```` as my keyboard does not help very much.

y```` - 16y = 0       

y=C(subscript 1)e^(2x)+C(subscript 2)e^(-2x)+C(subscript 3)sin(2x)+C(subscript 4)cos(2x)

Step 1-Find the fourth derivative

y`=2C(subscript 1)e^(2x)-2C(subscript 2)e^(-2x)+2C(subscript 3)cos(2x)-2C(subscript 4)sin(2x)

y``=4C(subscript 1)e^(2x)+4C(subscript 2)e^(-2x)-4C(subscript 3)sin(2x)-4C(subscript 4)cos(2x)

y```=8C(subscript 1)e^(2x)-8C(subscript 2)e^(-2x)-8C(subscript 3)cos(2x)+8C(subscript 4)sin(2x)

y````=16C(subscript 1)e^(2x)+16C(subscript 2)e^(-2x)+16C(subscript 3)sin(2x)+16C(subscript 4)cos(2x)

Step 2-Find -16y

-16(C[subscript 1]e^[2x]+C[subscript 2]e^[-2x]+C[subscript 3]sin[2x]+C[subscript 4]cos[2x]

-16C(subscript 1)e^(2x)-16C(subscript 2)e^(-2x)-16C(subscript 3)sin(2x)-16C(subscript 4)cos(2x)

Step 3-Plug and chug

16C(subscript 1)e^(2x)+16C(subscript 2)e^(-2x)+16C(subscript 3)sin(2x)+16C(subscript 4)cos(2x)-16C(subscript 1)e^(2x)-16C(subscript 2)e^(-2x)-16C(subscript 3)sin(2x)-16C(subscript 4)cos(2x)=0

Everything cancels...

0=0

So the function is a solution of the differential equation.

Ok...if it verifies the equation it means that it is a solution...then why did I end up with a contradiction?It may be faulty logic...