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 February 13th, 2014, 04:02 AM #1 Newbie   Joined: Feb 2014 Posts: 2 Thanks: 0 Elementary Differential Equation Problems Hello everyone, this is my first post, English is not my mother tongue, so i'm sorry if there's some wrong / unclear statement.. I was doing some exercise, the problem was y'' + 9y = 9 ( sec 3t )^2 ; 0 < t < pi/6 so the homogeneous solution would be = C1*cos 3t + C2*sin 3t and using Variation of Parameters i end up with the particular solution u1 =$$\int (\sec t) dt$$ u2 = $$\int (\sec t) (\tan t) dt$$ I don't know how to solve those integration or maybe the particular solution above was wrong cus of miscalculation please reply ASAP cus next week is my first exam
 February 13th, 2014, 05:18 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Elementary Differential Equation Problems I presume you know that $sec(t)= \frac{1}{cos(t)}$ and $tan(x)sec(x)= \frac{sin(x)}{cos(x)}\frac{1}{cos(x)}= \frac{sin(x)}{cos^2(x)}$ so your integrals are $\int \frac{1}{cos(t)}dt$ and $\int \frac{sin(t)}{cos^2(t)}dt$. Since the first has cos(t) to an odd power and the second has sin(t) to an odd power, there is a standard method of integration that you probably learned in Calculus. In the second you can immediately let u= cos(t) so that du= sin(t)dt and the integral becomes $\int \frac{du}{u}$. For the first, first multiply both numerator and denominator by cos(t) to get $\int \frac{cos(t)}{cos^2(t)}dt= \int \frac{cos(t)}{1- sin^2(t)}dt$. Now let u= sin(t) so that du= cos(t)dt and the integral becomes $\int\frac{du}{1- u^2}$ which can be done by "partial fractions".
 February 13th, 2014, 05:42 AM #3 Newbie   Joined: Feb 2014 Posts: 2 Thanks: 0 Re: Elementary Differential Equation Problems Oh My God I didn't Realize it was such a simple problem, Thank you so much
February 13th, 2014, 12:56 PM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

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Re: Elementary Differential Equation Problems

Hello, joseph95!

These are elementary integration problems.,
[color=beige]. . [/color]Did you forget the basic formulas?

Quote:
 $u_1 \;=\; \int \sec t\, dt$

$u_1 \;=\;\int \sec t\,dt \;=\;\ln|\sec t\,+\,\tan t| \,+\,C$

Quote:
 $u_2 \;=\; \int \sec t\,\!\tan t\,dt$

$u_2 \;=\;\int\sec t\,\!\tan t\,dt \;=\;\sec t\,+\,C$

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