How Do You Solve an Oscillating Body's Differential Equation?

[mxmadman_44]

Homework Statement

the deflection y of a body oscillating about a fixed reference point satisfies the ordinary differential equation

[math] d^2/dt^2 + 16y = 3sin(2t) [/math]

Where t is time

given the initial conditions y = 0 and dy/dt = 3/2 at time t= 0 , solve the differential equation to obtain an expression for the deflection y as a function of t.

Homework Equations

[math] d^2/dt^2 + 16y = 3sin(2t) [/math]

The Attempt at a Solution

I understand that i must let

[math] y = e^alphax [/math]

Therefor alpha^2 + 16 = 3sin(2t)

however i then know that the general solution is (x + 4)(x - 4)

however isn't this for a -16 not a positive 16?

but then i goto x^2 + 4x - 4x -16

Again -16?

but right there I am stuck

ive never solved one of these and I am at my limit with this one it really is beyond me.

Any help would be appreciated so much

Thanks

 

[rock.freak667]

so your equation is r²+16=0 to give you the homogeneous roots

use the quadratic equation formula and the fact that i=√-1 (imaginary unit) to get your roots in the form of λ±μi.

so y_h=e^λt(cos(μt)+sin(μt))

Now find y_p, then add that to y_h and you can go on.

 

[Mark44]

Let's look at the related, homogeneous diff. eqn: y'' + 16y = 0.
The characteristic equation is r² + 16 = 0, which has roots +/- 4i. This means that two solutions to the homogeneous diff. eqn. are e^^4t and e^-4t). The solutions to the homogeneous problem consist of all linear combinations of these two functions; i.e., the sum of constant multiples of these two functions.

For the nonhomogeneous diff. eqn. y'' + 16y = 3cos(2t), a good choice for a particular solution is y_p = Acos(2t). Substitute this function into your differential equation to solve for the constant A.

The general solution consists of the solutions to the homogeneous problem plus the particular solution.

 

[HallsofIvy]

Homework Statement

the deflection y of a body oscillating about a fixed reference point satisfies the ordinary differential equation

[math] d^2/dt^2 + 16y = 3sin(2t) [/math]

Where t is time

given the initial conditions y = 0 and dy/dt = 3/2 at time t= 0 , solve the differential equation to obtain an expression for the deflection y as a function of t.

Homework Equations

[math] d^2/dt^2 + 16y = 3sin(2t) [/math]

The Attempt at a Solution

I understand that i must let

[math] y = e^alphax [/math]

Therefor alpha^2 + 16 = 3sin(2t)

No, you mustn't. You must look at the associated homogeneous equation,
\frac{d^1y}{dt^2}+ 16= 0 which gives characteristic equation \alpha^2+ 16= 0

however i then know that the general solution is (x + 4)(x - 4)

No, you don't know that. In the first place, that is not a "solution" it is an attempt to factor. In the second place, its product is NOT "x^2+ 16", it is "x^2- 16" so your factoring is wrong. In the third place, it should be "\alpha", not x. If you write the equation as \alpha^2= -16 it should be easy to solve.

You will also want to use the fact that e^{i\alpha x}= cos(\alpha x)+ i sin(\alpha x)

however isn't this for a -16 not a positive 16?

but then i goto x^2 + 4x - 4x -16

Again -16?

but right there I am stuck

ive never solved one of these and I am at my limit with this one it really is beyond me.

Any help would be appreciated so much

Thanks

And, on this forum, use [ tex] or [ itex ] for LaTex.