MathSofiya
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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Algebraic!
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did you get that problem from the other day sorted out?
Algebraic!
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y" +y =t*e^t iirc?
MathSofiya
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at first I had r^2+r=0 and that gave me the integral of
\[\int\frac{e^{-x}sec^2x}{e^{-x}}dx+e^{-x}\int \frac{sec^2x}{e^{-x}}dx\]
MathSofiya
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but that is apparently wrong
MathSofiya
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iirc? what does that mean?
Algebraic!
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if I recall correctly
MathSofiya
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\[r_1=r_2=\pm1i\]
MathSofiya
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\[y_c=c_1cosx+c_2sinx\]
MathSofiya
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\[W(y_1,y_2)=cos^2x+sin^2x\]
Algebraic!
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1
MathSofiya
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nice!
\[y_p=-cosx\int sinxsec^2xdx+sinx \int cosxsec^2xdx\]
MathSofiya
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so far so good?
Algebraic!
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yes
MathSofiya
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\[y=y_c+y_p=c_1cosx+c_2sinx-cosxsecx+\int cosxsec^2xdx\]
Algebraic!
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missing a 'sin' on that last term...
psi9epsilon
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@mathsofiya , check integration seems incomplete
Algebraic!
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lol
psi9epsilon
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@Algebraic! , didnt really see the reason for "lol", there are couple of things missing here
Algebraic!
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"integration seems incomplete"
Algebraic!
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what tipped you off?
Algebraic!
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here I thought you were making a funny and you were somehow serious...
psi9epsilon
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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
: )
MathSofiya
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where's a sine missing?
Algebraic!
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sinx∫cosxsec^2x dx -> ∫cosxsec^2x dx
MathSofiya
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very true
\[y=y_c+y_p=c_1cosx+c_2sinx-cosxsecx+sinx\int cosxsec^2xdx\]
Algebraic!
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most people use ln(tan(x) +sec(x)) for brevity
Algebraic!
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programs always give the other form of the answer; not sure why, probably to confound students.