First order ODE? anyone good at them?

Question

d^2y/dx^2= - (k+1)^2 * y

Anyone know the general and particular solutions for this?

Answer 1

1st off this is a 2nd order DE...the solution should be obvious..y ' ' + (k+1)² y = 0 means trig functions,,,,c1 cos (k+1) x + c2 sin(k+1) x

Answer 2

the k+1 part throws me off. what the heck are u supposed to do with k+1^2

...Ron shouldnt the soln be e[Acos(k+1)x+Bsin(k+1)x]? I am confused. I did what u guys said above and got that as an answer

Nevermind it is Acos(k+1)x+Bsin(k+1)x...You guys are smart

Answer 3

It's second order, not first order.

The general solution is

y = c1 cos[(k+1)x] + c2 sin[(k+1)x]

where c1 and c2 are arbitrary constants.

This may be established by trying a solution of the form

y = e^(mx)

For this equation, we get m² + (k+1)² = 0, whose solutions are m = ±i(k+1) where i is the imaginary unit (√(-1)) Using e^(iθ) = cos(θ) + i sin(θ) and the fact that the real and imaginary parts must each satisfy the ODE gives us the cosine and sine solutions shown above.