Auxiliary Eqns with Complex Roots

[marco101]

Homework Statement

solve the initial value problem
y''+9y = 0, y(0) = 1, y'(0) = 1

Homework Equations

gen solution form is y(t) = C₁e^At*(cosBt) + C₂e^At*(sinBt)
where A is the real number and B is the imaginary number

The Attempt at a Solution

i just wanted to check if I am doing this right.
i started off finding the roots,
r^2 + 9
r = +/- 3i
since the real is 0, e would be raised to 0 and become one leaving me with C₁cos3t + C₂sin3t

 

[rock.freak667]

The Attempt at a Solution

i just wanted to check if I am doing this right.
i started off finding the roots,
r^2 + 9
r = +/- 3i
since the real is 0, e would be raised to 0 and become one leaving me with C₁cos3t + C₂sin3t

Yes that is correct. Now find y(0) and y'(0)

 

[marco101]

since there are double roots, should i make changes to the general solution form? everything cancels out to be zero because y'' = -9cos3t -9sin3t and y = 9cos3t + 9sin3t because it was multiplied by 9 in the original equation.

 

[rock.freak667]

since there are double roots, should i make changes to the general solution form? everything cancels out to be zero because y'' = -9cos3t -9sin3t and y = 9cos3t + 9sin3t because it was multiplied by 9 in the original equation.

double roots would mean r=3i,3i or r=-3i,-3i. You have r=0±3i. So yes your general solution is correct.

Also from y = 9cos3t + 9sin3t, y''≠ y'' = -9cos3t -9sin3t , recheck your algebra on that one.

Anyway you take it y= 9cos3t +9sin3t is correct.