Question

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

Answer 1

did you get that problem from the other day sorted out?

Answer 2

y" +y =t*e^t iirc?

Answer 3

at first I had r^2+r=0 and that gave me the integral of
\[\int\frac{e^{-x}sec^2x}{e^{-x}}dx+e^{-x}\int \frac{sec^2x}{e^{-x}}dx\]

Answer 4

but that is apparently wrong

Answer 5

iirc? what does that mean?

Answer 6

if I recall correctly

Answer 7

\[r_1=r_2=\pm1i\]

Answer 8

\[y_c=c_1cosx+c_2sinx\]

Answer 9

\[W(y_1,y_2)=cos^2x+sin^2x\]

Answer 10

1

Answer 11

nice!
\[y_p=-cosx\int sinxsec^2xdx+sinx \int cosxsec^2xdx\]

Answer 12

so far so good?

Answer 13

yes

Answer 14

\[y=y_c+y_p=c_1cosx+c_2sinx-cosxsecx+\int cosxsec^2xdx\]

Answer 15

missing a 'sin' on that last term...

Answer 16

@mathsofiya , check integration seems incomplete

Answer 17

lol

Answer 18

@Algebraic! , didnt really see the reason for "lol", there are couple of things missing here

Answer 19

"integration seems incomplete"

Answer 20

what tipped you off?

Answer 21

here I thought you were making a funny and you were somehow serious...

Answer 22

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

: )

Answer 23

where's a sine missing?

Answer 24

sinx∫cosxsec^2x dx -> ∫cosxsec^2x dx

Answer 25

very true
\[y=y_c+y_p=c_1cosx+c_2sinx-cosxsecx+sinx\int cosxsec^2xdx\]

Answer 26

is this really the solution to the integral?
http://www.wolframalpha.com/input/?i=integral+of+cosxsec^2x

Answer 27

most people use ln(tan(x) +sec(x)) for brevity

Answer 28

programs always give the other form of the answer; not sure why, probably to confound students.