# Variation of Parameters

• Apr 10th 2011, 08:57 PM
Naples
Variation of Parameters
Solve using Variation of Parameters

1. $\displaystyle y" + y = tan(x)$
auxiliary equation
$\displaystyle m^2 + 1 = 0$
$\displaystyle Yc = c*cos(x) + c*sin(x)$

$\displaystyle Y1 = cos(x)$
$\displaystyle Y2 = sin(x)$
Wronskian = 1
$\displaystyle f(x) = tan(x)$

$\displaystyle u1' = -sin(x)tan(x)$
$\displaystyle u1 = ?$ Not sure how to integrate this...

$\displaystyle u2' = cos(x)tan(x) = sin(x)$
$\displaystyle u2 = -cos(x)$

I'd appreciate it if someone could check over what I've done so far and then tell me how to integrate $\displaystyle -sin(x)tan(x)$.

2.
$\displaystyle y" - 16y = 2e^4^x$
auxiliary equation
$\displaystyle m^2 - 16 = 0$
$\displaystyle Yc = ce^-^4^x + ce^4^x$

$\displaystyle Y1 = e^-4^x$
$\displaystyle Y2 = e^4^x$
Wronskian = 8
$\displaystyle f(x) = 2e^4^x$

$\displaystyle u1' = -(1/4)e^8^x$
$\displaystyle u1 = -(1/32)e^8^x$

$\displaystyle u2' = 1/4$
$\displaystyle u2 = (1/4)x$

$\displaystyle Yp = (1/4)xe^4^x - (1/32)e^4^x$
so $\displaystyle Y = ce^-^4^x + ce^4^x + (1/4)xe^4^x - (1/32)e^4^x$

Not sure if I did something wrong because when I used undetermined coefficients to solve the problem, I only got $\displaystyle Yp = (1/4)x*e^4^x$...?

• Apr 10th 2011, 09:03 PM
Chris L T521
Quote:

Originally Posted by Naples
Solve using Variation of Parameters

1. $\displaystyle y" + y = tan(x)$
auxiliary equation
$\displaystyle m^2 + 1 = 0$
$\displaystyle Yc = c*cos(x) + c*sin(x)$

$\displaystyle Y1 = cos(x)$
$\displaystyle Y2 = sin(x)$
Wronskian = 1
$\displaystyle f(x) = tan(x)$

$\displaystyle u1' = -sin(x)tan(x)$
$\displaystyle u1 = ?$ Not sure how to integrate this...

$\displaystyle u2' = cos(x)tan(x) = sin(x)$
$\displaystyle u2 = -cos(x)$

I'd appreciate it if someone could check over what I've done so far and then tell me how to integrate $\displaystyle -sin(x)tan(x)$.

Spoiler:
$\displaystyle -\sin x\tan x = -\dfrac{\sin^2x}{\cos x} = \cos x-\sec x$

Quote:

2.
$\displaystyle y" - 16y = 2e^4^x$
auxiliary equation
$\displaystyle m^2 - 16 = 0$
$\displaystyle Yc = ce^-^4^x + ce^4^x$

$\displaystyle Y1 = e^-4^x$
$\displaystyle Y2 = e^4^x$
Wronskian = 8
$\displaystyle f(x) = 2e^4^x$

$\displaystyle u1' = -(1/4)e^8^x$
$\displaystyle u1 = -(1/32)e^8^x$

$\displaystyle u2' = 1/4$
$\displaystyle u2 = (1/4)x$

$\displaystyle Yp = (1/4)xe^4^x - (1/32)e^4^x$
so $\displaystyle Y = ce^-^4^x + ce^4^x + (1/4)xe^4^x - (1/32)e^4^x$

Not sure if I did something wrong because when I used undetermined coefficients to solve the problem, I only got $\displaystyle Yp = (1/4)x*e^4^x$...?

It looks good. Note that we can say that $\displaystyle ce^{4x}-\frac{1}{32}e^{4x} \sim ke^{4x};\,\,k=c-\frac{1}{32}$, so it doesn't really add "new" information.

I hope this makes sense.
• Apr 11th 2011, 07:52 AM
topsquark
Quote:

Originally Posted by Naples
1. $\displaystyle y" + y = tan(x)$
auxiliary equation
$\displaystyle m^2 + 1 = 0$
$\displaystyle Yc = c*cos(x) + c*sin(x)$

Just a quick little comment that has nothing to do with variation of parameters. The homogeneous solution to this equation is
$\displaystyle Yc = c_1*cos(x) + c_2*sin(x)$

ie the arbitrary constants are not the same. You did this in your second example as well.

-Dan