I'm trying to find the solution to ${y}''=-9y$

I've worked out that ${y}=\cos(-3x)+\sin(-3x)$ seems to fit, but the book says that the initial conditions should be ${y}(0)=1$ and ${y}'(0)=0$

But when I plug $0$ into this function I don't get $0$ ${y}'(0)= 3\sin(-3x)-3\cos(-3x) = -3$

Where am I going wrong?


Here is a full solution for reference.

Note that the characteristic equation is $r^2+9=0$. The solutions to this are $r=0\pm3i$, so we know that the general solution to our differential equation is in the form


for some real constants $c_1$ and $c_2$.

Now we can plug in our initial conditions to find our solution. The initial condition $1=y(0)$ gives us


while the initial condition $0=y'(0)$ gives us


Hence $c_1=1$ and $c_2=0$. This gives us our final solution of



You forgot the constants..the general solution is of the form: $$ y = A\sin(3x) + B\cos(3x) $$ So try to pick constants $A$ and $B$ to fit the conditions.

Your solution is saying that $A=B=1$ but your initial conditions are saying otherwise.

First bit $$ y(0) = A\sin(3\cdot 0) + B \cos(3\cdot 0)= B $$ What must $B$ be to satisfy the condition on $y$?


Oops I forgot to mention you also got the argument of your sines and cosines wrong it should not be $-3x$ a common mistake when people come across these Simple hamonic motion equations.

  • 1
    $\begingroup$ I don't think the argument of the sines and cosines matters since we have $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$ - in essence all it will do is change the sign of your $A$ constant (though probably things will look prettier if there is no "$-$" sign in the arguments). $\endgroup$ – Peter Woolfitt Dec 31 '14 at 10:35
  • $\begingroup$ You are right. I just a little biased with the way I was taught. Though I am laughing at myself now since I have never thought about it like that .. With using the property of the odd parity with the sines.. $\endgroup$ – Chinny84 Dec 31 '14 at 10:39

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