- #1

BillhB

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## Homework Statement

Use variation of parameters to solve ##\frac{d^2y}{dx^2} + 4y = sec(2x)##

## Homework Equations

Really just curious about comments, see picture.

## The Attempt at a Solution

I'm having a hard time deciphering the comments I got back on an exam.

For the first comment, is there a system of equations? Sure, they were:

$$u_1'y_1+u_2'y_2=0$$

$$u_1'y_1'+u_2'y_2'=g(t)$$

I don't see the point of writing that on an exam, where there's seven other questions with a hour time limit.

He circled the Wronskian and asked what it represented... the determinant of the Wronskian ?##y_1y_2'-y_2y_1'= W##

The note about it not being variation of parameters is the most confusing... what do you call it then?

Then there's the note about where does this come from? I guess I could re-derive the general formula, but I didn't feel it was necessary at the time. Lesson learned.

The next comment confuses me as well, how are:

$$y_p = -cos(2x)\int{\frac{tan(2x)dx}{2}}+sin(2x)\int{\frac{dx}{2}}$$

$$y_p=-cos(2x)\int{\frac{-du}{4u}} + \frac{1}{2}xsin(2x)+C_2$$

$$y_p= \frac{1}{4}cos(2x)ln|cos(2x)| +C_1 + \frac{1}{2}xsin(2x)+C_2$$

Not equal, just not seeing it.

I guess I should have just wrote...

$$y=\frac{1}{4}cos(2x)(ln|cos(2x)|+C_1) + \frac{1}{2}sin(2x)(x+C_2)$$

And that would just give the general solution. Probably better that way instead of making ##C_1 = C_2 = 0## in my head and getting sloppy to keep it as ##y_p##

On the bit about there being no arbitrary constants in the particular solution, sure, probably didn't handle that the best way. The book says treat them as zero. I'm thinking maybe just taking definite integrals from 0 to x would be a better way to say the same thing or just changing ##y_p## to ##y##

The comment about how many answers? There's infinity many. The constants in front of the homogeneous part can be whatever. Not understanding how I missed 90% of the solution.

I talked to some other classmates, they said I HAD to do it using Cramer’s . Uhhhh...? Why? It leads to the same thing, especially since the order was only two.

Thanks for any help or advice!

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