1. ## Variation of Parameters

Solve using Variation of Parameters

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

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

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

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

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

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

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

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

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

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

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

2. Originally Posted by Naples
Solve using Variation of Parameters

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

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

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

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

I'd appreciate it if someone could check over what I've done so far and then tell me how to integrate $-sin(x)tan(x)$.
Spoiler:
$-\sin x\tan x = -\dfrac{\sin^2x}{\cos x} = \cos x-\sec x$

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

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

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

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

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

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

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

I hope this makes sense.

3. Originally Posted by Naples
1. $y" + y = tan(x)$
auxiliary equation
$m^2 + 1 = 0$
$Yc = c*cos(x) + c*sin(x)$
Just a quick little comment that has nothing to do with variation of parameters. The homogeneous solution to this equation is
$Yc = c_1*cos(x) + c_2*sin(x)$

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

-Dan

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### y"-16y=2e^4x

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