1. ## another differential equation

i need help with this one
solve
$y'' + 9y = 0; \quad y = 3, \; y'=3 \;\textnormal{at}\; x = \pi/3$

$y(x) = ?$

2. Write down the characteristic equation. Then see what form the solution has. So $r^2 + 9r = 0$ or $r(r+9) = 0$, which means $r = 0, - 9$ which are real solutions. Then the general solution has the form $y = c_{1}e^{r_{1}t} + c_{2}e^{r_{2}t}$. So $y' = c_{1}r_{1}e^{r_{1}t} + c_{2}r_{2}e^{r_{2}t}$. You are given the initial conditions.

3. Originally Posted by tukeywilliams
Write down the characteristic equation. Then see what form the solution has. So $r^2 + 9r = 0$ or $r(r+9) = 0$, which means $r = 0, - 9$ which are real solutions. Then the general solution has the form $y = c_{1}e^{r_{1}t} + c_{2}e^{r_{2}t}$. So $y' = c_{1}r_{1}e^{r_{1}t} + c_{2}r_{2}e^{r_{2}t}$. You are given the initial conditions.
The characteristic equation here is $r^2+9=0$, so $r=\pm ~ 3 \bold{i}$

RonL

4. Whoops. Sorry about that. In that case the general solution would be $y(t) = c_{1}e^{(a+\bold{i}b)t} + c_{2}e^{(a - \bold{i}b)t}$.

5. When $r=a\pm bi$, the general solution is
$y=e^{ax}(C_1\cos bx+C_2\sin bx)$

6. Originally Posted by tukeywilliams
Whoops. Sorry about that. In that case the general solution would be $y(t) = c_{1}e^{(a+\bold{i}b)t} + c_{2}e^{(a - \bold{i}b)t}$.
Originally Posted by red_dog
When $r=a\pm bi$, the general solution is
$y=e^{ax}(C_1\cos bx+C_2\sin bx)$
These are the same thing in different notation.

RonL

7. im a little confused on applying the conditions

8. $y(x)=C_1\cos 3x+C_2\sin 3x$
$y'(x)=-3C_1\sin 3x+3C_2\cos 3x$
Now, put $x=\frac{\pi}{3}$ and solve the system
$\left\{\begin{array}{ll}y\left(\frac{\pi}{3}\right )=3\\y \ '\left(\frac{\pi}{3}\right)=3\end{array}\right.$