Finding Particle Paths: Solving a Complimentary Function

[maggie56]

I don't understand how to find particle paths, for example i have a question that states;

u= (-z + cos(at)) j + (y + sin(at)) k

for the complementary function

y' = -z
x' = y

so y''=-y therefore y = A cos t + B sin t and z = A sin t - B cos t

Now for the particular integral, i know the answer is
y=1/(a-1) sin (at) and z = -1/(a-1) cos (at)
i assume this has been found using a linear combination of cos at and sin at but i don't see how

Could someone please help

Thanks

 

[vela]

Could you state the problem as it was given to you word for word? You seem to be leaving out some critical information.

 

[maggie56]

It just says what is the particle path of the flow u= (-z + cos(at)) j + (y + sin(at)) k
It is an example from a lecture, previously we had found the streamlines for the flow.

 

[vela]

You can use the method of undetermined coefficients: Write the particular solutions as

y_p = c₁ cos at + d₁ sin at
z_p = c₂ cos at + d₂ sin at

and plug them into the equations

y' = -z + cos at
z' = y + sin at (I'm assuming here there was a typo in your original post, x' instead of z'.)

Match the coefficients of cos at and sin at to end up with four equations which you can solve to find the constants.

 

[HallsofIvy]

The "critical information" you were leaving out in your first post was that u is the velocity vector at point (x, y, z) and time t.