Variation of Parameters second order ODE
$begingroup$
Sorry in advance for the chaotic mess. Hopefully you will be able to follow. Your help will be greatly appreciated.
Given the second order ODEP: $y" + 9y = 6tan(3x)$
Find the characteristic polynomial:
$r = 0 pm 3i$
$y_1= cos(3x)$
$y_2= sine(3x)$
$W(y_1,y_2)=3$
$-cos(3x) int frac{sine(3x)6sin(3x)}{3cos(3x)} + sine(3x) int frac{6cos(3x)sin(3x)}{3cos(3x)}$
I am unsure as to why The LHS goes from $frac{1-cos^2(3x)}{cos(3x)}$
to $-cos(3x) int sec(3x) - cos(3x)+ sin(3x) int ...$
is the because the integral could be rewritten as $int frac{1}{cos(3x)} times frac{-cos^2(3x)}{cos(3x)}$ since $frac{1}{cos(x)} = sec(3x)$
Then where does everything come from after that in the following, besides the natural log part. That is straight forward.
$frac{-cos(3t)}{3}(ln vert sec(3x) + tan(3x) vert - sin(3x)$
$-sin(3t)$ and the three in the denominator of $cos(3x)$?
Why is the 6 not included in the integrals as G(x)?
and finally I don't understand where $int sin(3x)$ goes to
$frac{sin(3x)}{3} times -cos(3x)$?
calculus ordinary-differential-equations
asked Jan 8 at 23:00
Austin DennyAustin Denny
61
$endgroup$
add a comment |
$begingroup$
Sorry in advance for the chaotic mess. Hopefully you will be able to follow. Your help will be greatly appreciated.
Given the second order ODEP: $y" + 9y = 6tan(3x)$
Find the characteristic polynomial:
$r = 0 pm 3i$
$y_1= cos(3x)$
$y_2= sine(3x)$
$W(y_1,y_2)=3$
$-cos(3x) int frac{sine(3x)6sin(3x)}{3cos(3x)} + sine(3x) int frac{6cos(3x)sin(3x)}{3cos(3x)}$
I am unsure as to why The LHS goes from $frac{1-cos^2(3x)}{cos(3x)}$
to $-cos(3x) int sec(3x) - cos(3x)+ sin(3x) int ...$
is the because the integral could be rewritten as $int frac{1}{cos(3x)} times frac{-cos^2(3x)}{cos(3x)}$ since $frac{1}{cos(x)} = sec(3x)$
Then where does everything come from after that in the following, besides the natural log part. That is straight forward.
$frac{-cos(3t)}{3}(ln vert sec(3x) + tan(3x) vert - sin(3x)$
$-sin(3t)$ and the three in the denominator of $cos(3x)$?
Why is the 6 not included in the integrals as G(x)?
and finally I don't understand where $int sin(3x)$ goes to
$frac{sin(3x)}{3} times -cos(3x)$?
calculus ordinary-differential-equations
asked Jan 8 at 23:00
Austin DennyAustin Denny
61
$endgroup$
add a comment |
$begingroup$
Sorry in advance for the chaotic mess. Hopefully you will be able to follow. Your help will be greatly appreciated.
Given the second order ODEP: $y" + 9y = 6tan(3x)$
Find the characteristic polynomial:
$r = 0 pm 3i$
$y_1= cos(3x)$
$y_2= sine(3x)$
$W(y_1,y_2)=3$
$-cos(3x) int frac{sine(3x)6sin(3x)}{3cos(3x)} + sine(3x) int frac{6cos(3x)sin(3x)}{3cos(3x)}$
I am unsure as to why The LHS goes from $frac{1-cos^2(3x)}{cos(3x)}$
to $-cos(3x) int sec(3x) - cos(3x)+ sin(3x) int ...$
is the because the integral could be rewritten as $int frac{1}{cos(3x)} times frac{-cos^2(3x)}{cos(3x)}$ since $frac{1}{cos(x)} = sec(3x)$
Then where does everything come from after that in the following, besides the natural log part. That is straight forward.
$frac{-cos(3t)}{3}(ln vert sec(3x) + tan(3x) vert - sin(3x)$
$-sin(3t)$ and the three in the denominator of $cos(3x)$?
Why is the 6 not included in the integrals as G(x)?
and finally I don't understand where $int sin(3x)$ goes to
$frac{sin(3x)}{3} times -cos(3x)$?
calculus ordinary-differential-equations
asked Jan 8 at 23:00
Austin DennyAustin Denny
61
$endgroup$
Sorry in advance for the chaotic mess. Hopefully you will be able to follow. Your help will be greatly appreciated.
Given the second order ODEP: $y" + 9y = 6tan(3x)$
Find the characteristic polynomial:
$r = 0 pm 3i$
$y_1= cos(3x)$
$y_2= sine(3x)$
$W(y_1,y_2)=3$
$-cos(3x) int frac{sine(3x)6sin(3x)}{3cos(3x)} + sine(3x) int frac{6cos(3x)sin(3x)}{3cos(3x)}$
I am unsure as to why The LHS goes from $frac{1-cos^2(3x)}{cos(3x)}$
to $-cos(3x) int sec(3x) - cos(3x)+ sin(3x) int ...$
is the because the integral could be rewritten as $int frac{1}{cos(3x)} times frac{-cos^2(3x)}{cos(3x)}$ since $frac{1}{cos(x)} = sec(3x)$
Then where does everything come from after that in the following, besides the natural log part. That is straight forward.
$frac{-cos(3t)}{3}(ln vert sec(3x) + tan(3x) vert - sin(3x)$
$-sin(3t)$ and the three in the denominator of $cos(3x)$?
Why is the 6 not included in the integrals as G(x)?
and finally I don't understand where $int sin(3x)$ goes to
$frac{sin(3x)}{3} times -cos(3x)$?
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Jan 8 at 23:00
Austin DennyAustin Denny
61
asked Jan 8 at 23:00
Austin DennyAustin Denny
61
asked Jan 8 at 23:00
Austin DennyAustin Denny
61
asked Jan 8 at 23:00
Austin DennyAustin Denny
61
asked Jan 8 at 23:00
Austin DennyAustin Denny
61
61
add a comment |
add a comment |
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