Is this 2nd Differential Equation Answer Correct?

[jonnejon]

Homework Statement

y'' + 9y = 0 where y(pi/3) = y'(pi/3) = 3

Homework Equations

r^2 + 9 = 0
r^2 = -9
r = 3i

The Attempt at a Solution

y = c1 e^x cos (3x) + c2 e^x sin(3x)
y' = c1 e^x cos (3x) - 3c1 e^x sin (3x) + c2 e^x sin(3x) + 3c2 e^x cos(3x)

y = -c1 e^(pi/3) = 3 => c1 = -3e^(-pi/3)
y' = -c1 e^(pi/3) - 3c2 e^(pi/3) = 3
y' = -(-3e^(-pi/3))e^(pi/3) - 3c2 e^(pi/3) = 3
y' = 3 - 3c2e^(pi/3) = 3 ====> c2 = 0

so
y= -3e^(-pi/3) e^x cos(3x) + 0

 

[tiny-tim]

Hi jonnejon!

(have a pi: π and try using the X² and X₂ tags just above the Reply box )

r = 3i
y = c1 e^x cos (3x) + c2 e^x sin(3x)

Noooo … r = ±3i gives you Ae^3ix + Be^-3ix (or Acos3x + Bsin3x).

(Your solution would be for r = 1 ± 3i. )

 

[jonnejon]

So,

y = A cos(3x) + B sin(3x)
y' = -3A sin(3x) + 3B sin(3x)

y = -A = 3 => A = -3
y'= -3B = 3 => B = -1

So the solution is: y= 3cos(3x) - sin(3x)??

 

[tiny-tim]

So,

y = A cos(3x) + B sin(3x)
y' = -3A sin(3x) + 3B sin(3x)

y = -A = 3 => A = -3
y'= -3B = 3 => B = -1

So the solution is: y= 3cos(3x) - sin(3x)??

Yup!

(except you left out the - in -3 )

 

[Mark44]

So,

y = A cos(3x) + B sin(3x)
y' = -3A sin(3x) + 3B sin(3x)

For y' it should by y' = -3A sin(3x) + 3B cos(3x)
If I can jump in here, the next two lines are where you're using your initial conditions to solve for A and B in the equations above.

y = -A = 3 => A = -3
y'= -3B = 3 => B = -1

What you should say in the first equation is that since y(pi/3) = 3, then
Acos(pi) + Bsin(pi) = 3, so A*(-1) = 3, or A = -3

Since y'(pi/3) = 3, and A = -3,
9sin(pi) + 3B cos(pi) = 3, so 3B(-1) = 3, so B = -1

So the solution is: y= 3cos(3x) - sin(3x)??

As tiny-tim already pointed out, y = -3cos(3x) - sin(3x)

 

[tiny-tim]

As Compuchip already pointed out, y = -3cos(3x) - sin(3x)

Hi Compuchip!

Didn't know you were there! ​

 

[Mark44]

Sorry about that, tiny-tim.

 

[tiny-tim]

We look alike in the dark!

 

[Mark44]

Well, it's dark here, so that's good enough for me.