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anonymous
 3 years ago
solve the differential equation using the variation of parameters method
\[y''+y=sec^2x,0\le x\le \frac{\pi}2\]
\[r^2+1=0\]
anonymous
 3 years ago
solve the differential equation using the variation of parameters method \[y''+y=sec^2x,0\le x\le \frac{\pi}2\] \[r^2+1=0\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you get that problem from the other day sorted out?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0at first I had r^2+r=0 and that gave me the integral of \[\int\frac{e^{x}sec^2x}{e^{x}}dx+e^{x}\int \frac{sec^2x}{e^{x}}dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but that is apparently wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0iirc? what does that mean?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if I recall correctly

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y_c=c_1cosx+c_2sinx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[W(y_1,y_2)=cos^2x+sin^2x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nice! \[y_p=cosx\int sinxsec^2xdx+sinx \int cosxsec^2xdx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y=y_c+y_p=c_1cosx+c_2sinxcosxsecx+\int cosxsec^2xdx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0missing a 'sin' on that last term...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mathsofiya , check integration seems incomplete

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! , didnt really see the reason for "lol", there are couple of things missing here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"integration seems incomplete"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here I thought you were making a funny and you were somehow serious...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a 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
 3 years ago
Best ResponseYou've already chosen the best response.0where's a sine missing?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sinx∫cosxsec^2x dx > ∫cosxsec^2x dx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0very true \[y=y_c+y_p=c_1cosx+c_2sinxcosxsecx+sinx\int cosxsec^2xdx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is this really the solution to the integral? http://www.wolframalpha.com/input/?i=integral+of+cosxsec^2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0most people use ln(tan(x) +sec(x)) for brevity

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0programs always give the other form of the answer; not sure why, probably to confound students.
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