# Math Help - Differential Equations and Variation of Parameters

1. For the last solution,

y'''+y'=sec t

You can let z=y'

To get,

z''+z=sec t

The general solution to homogenous equation is:
z=C_1sin(t)+C_2cos(t)

Then you can find particular via Lagrange's Variation of Parameters techiqnue.

Then you need to integrate the entire solution for z tto get y.
Couldnt you let L(y) = y''' + p(t)y'' + q(t)y' + r(t)y = f(t)

and y_h = c_1y_! + c_2y_2 + c_3y_3 and y_p = v_1y_1+v_2y_2+v_3y_3

Then v_1', v_2, and v'_3 have to satisfy some auxillary condition, and you can use Crameris Rule to get the solution.

2. Originally Posted by tukeywilliams
Couldnt you let L(y) = y''' + p(t)y'' + q(t)y' + r(t)y = f(t)

and y_h = c_1y_! + c_2y_2 + c_3y_3 and y_p = v_1y_1+v_2y_2+v_3y_3

Then v_1', v_2, and v'_3 have to satisfy some auxillary condition, and you can use Crameris Rule to get the solution.
I am not sure what you are doing.

There is a Generalized Variation of Parameters.

But again it involves finding all 3 independent solution to,
y'''+y'=0
Which is,
z''+z=0 (after reduction of order).
Hence,
z=y'=C_1sin(t)+C_2cos(t)
But then,
y=C_1sin(t)+C_2cos(t)+C_3

Thus,
y_1=sin(t)
y_2=cos(t)
y_3=1
Are three linearly independent solutions.

Now! You can use the generalized variation of parameters techinique for linear differencial equations of order 3.

3. Heres what I did roughly:

Here

4. Originally Posted by tukeywilliams
Heres what I did roughly:

Here
It looks cool, but I have no idea what you did.

5. I edited it:

Here

6. Originally Posted by tukeywilliams
I edited it:

Here
That looks right.
W is the Wronskian (which you know because of my earlier post).

I see what you are saying.
That is the standard way I believe multi-dimensional variation of parameters is set up. However, that is not the solution. You need to painfully evaluate each determinant and them integrate all three v_1',v_2',v_3'.

It takes a longgg time to do that.

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By the Way: You write like a girl!

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