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

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

but that is apparently wrong

5. anonymous

iirc? what does that mean?

6. anonymous

if I recall correctly

7. anonymous

$r_1=r_2=\pm1i$

8. anonymous

$y_c=c_1cosx+c_2sinx$

9. anonymous

$W(y_1,y_2)=cos^2x+sin^2x$

10. anonymous

1

11. anonymous

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

12. anonymous

so far so good?

13. anonymous

yes

14. anonymous

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

15. anonymous

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

16. anonymous

@mathsofiya , check integration seems incomplete

17. anonymous

lol

18. anonymous

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

19. anonymous

"integration seems incomplete"

20. anonymous

what tipped you off?

21. anonymous

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

22. 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 : )

23. anonymous

where's a sine missing?

24. anonymous

sinx∫cosxsec^2x dx -> ∫cosxsec^2x dx

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

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