write the general solution to the differential equation

**slapmaxwell1** · March 26th 2011, 09:46 PM

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

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

is this right?

**Prove It** · March 26th 2011, 09:53 PM

I assume this is supposed to mean

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

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

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

a second order linear constant coefficient homogeneous ordinary differential equation.

I'm sure you can solve for $\displaystyle Y = \frac{d^2y}{dx^2}$ , and use this to solve for $\displaystyle y$ .

**slapmaxwell1** · March 26th 2011, 10:06 PM

here is a problem i did earlier and got the answer right maybe this example can show what im trying to do...

$D(D+9)(D-6)y=0\\ y=C_{1}+C_{2}e^{-2x}+C_{3}e^{6x}$

**slapmaxwell1** · March 26th 2011, 10:07 PM

sorry here is the correction..

$D(D+2)(D-6)y=0\\ y=C_{1}+C_{2}e^{-2x}+C_{3}e^{6x}$

**HallsofIvy** · March 27th 2011, 03:39 AM

Originally Posted by slapmaxwell1

sorry here is the correction..

$D(D+2)(D-6)y=0\\ y=C_{1}+C_{2}e^{-2x}+C_{3}e^{6x}$

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 $D^2(D^2+ 16)= 0$ 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 $Ae^{0x}= A$ and $Bxe^{0x}= Bx$ . The general solution to the equation is $y(x)= A+ Bx+ Ccos(4x)+ Dsin(4x)$ .

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