D^2(D^2+16)y=0

ok so this is y=xC1 + C2sin 4x + C3cos 4x

is this right?

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- Mar 26th 2011, 09:46 PMslapmaxwell1write the general solution to the differential equation
D^2(D^2+16)y=0

ok so this is y=xC1 + C2sin 4x + C3cos 4x

is this right? - Mar 26th 2011, 09:53 PMProve It
I assume this is supposed to mean

$\displaystyle \displaystyle \frac{d^4y}{dx^2} + 16\frac{d^2y}{dx^2} = 0$.

If you make the substitution $\displaystyle \displaystyle Y = \frac{d^2y}{dx^2}$, then this DE becomes

$\displaystyle \displaystyle \frac{d^2Y}{dx^2} + 16Y = 0$,

a second order linear constant coefficient homogeneous ordinary differential equation.

I'm sure you can solve for $\displaystyle \displaystyle Y = \frac{d^2y}{dx^2}$, and use this to solve for $\displaystyle \displaystyle y$. - Mar 26th 2011, 10:06 PMslapmaxwell1
here is a problem i did earlier and got the answer right maybe this example can show what im trying to do...

http://latex.codecogs.com/gif.latex?...x}+C_{3}e^{6x} - Mar 26th 2011, 10:07 PMslapmaxwell1
sorry here is the correction..

http://latex.codecogs.com/gif.latex?...x}+C_{3}e^{6x} - Mar 27th 2011, 03:39 AMHallsofIvy
Yes, that is correct. The characteristic equation can be written D(D+2)(D- 6)= 0 wo D= 0, -2, and 6 are characteristic roots.

For your first problem, that's a fourth order equation and must have 4 indpendent solutions. The characteristic equation was $\displaystyle D^2(D^2+ 16)= 0$ which has D= 0 as a double root and D= 4i and D= -4i as the other roots. Because of the double root, you need both $\displaystyle Ae^{0x}= A$ and $\displaystyle Bxe^{0x}= Bx$. The general solution to the equation is $\displaystyle y(x)= A+ Bx+ Ccos(4x)+ Dsin(4x)$.