anonymous
  • anonymous
solve the differential equation using the variation of parameters method \[y''+y=sec^2x,0\le x\le \frac{\pi}2\] \[r^2+1=0\]
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
did you get that problem from the other day sorted out?
anonymous
  • anonymous
y" +y =t*e^t iirc?
anonymous
  • anonymous
at first I had r^2+r=0 and that gave me the integral of \[\int\frac{e^{-x}sec^2x}{e^{-x}}dx+e^{-x}\int \frac{sec^2x}{e^{-x}}dx\]

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anonymous
  • anonymous
but that is apparently wrong
anonymous
  • anonymous
iirc? what does that mean?
anonymous
  • anonymous
if I recall correctly
anonymous
  • anonymous
\[r_1=r_2=\pm1i\]
anonymous
  • anonymous
\[y_c=c_1cosx+c_2sinx\]
anonymous
  • anonymous
\[W(y_1,y_2)=cos^2x+sin^2x\]
anonymous
  • anonymous
1
anonymous
  • anonymous
nice! \[y_p=-cosx\int sinxsec^2xdx+sinx \int cosxsec^2xdx\]
anonymous
  • anonymous
so far so good?
anonymous
  • anonymous
yes
anonymous
  • anonymous
\[y=y_c+y_p=c_1cosx+c_2sinx-cosxsecx+\int cosxsec^2xdx\]
anonymous
  • anonymous
missing a 'sin' on that last term...
anonymous
  • anonymous
@mathsofiya , check integration seems incomplete
anonymous
  • anonymous
lol
anonymous
  • anonymous
@Algebraic! , didnt really see the reason for "lol", there are couple of things missing here
anonymous
  • anonymous
"integration seems incomplete"
anonymous
  • anonymous
what tipped you off?
anonymous
  • anonymous
here I thought you were making a funny and you were somehow serious...
anonymous
  • anonymous
a simple look suggests that last term carries an Integration sign whereas the second last term does not, it cant be both , either we have an integral sign or we dont. Anyways its better to teach something correctly : )
anonymous
  • anonymous
where's a sine missing?
anonymous
  • anonymous
sinx∫cosxsec^2x dx -> ∫cosxsec^2x dx
anonymous
  • anonymous
very true \[y=y_c+y_p=c_1cosx+c_2sinx-cosxsecx+sinx\int cosxsec^2xdx\]
anonymous
  • anonymous
is this really the solution to the integral? http://www.wolframalpha.com/input/?i=integral+of+cosxsec^2x
anonymous
  • anonymous
most people use ln(tan(x) +sec(x)) for brevity
anonymous
  • anonymous
programs always give the other form of the answer; not sure why, probably to confound students.

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