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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
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did you get that problem from the other day sorted out?
y" +y =t*e^t iirc?
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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Other answers:

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

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