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1. ### Find the solution of the following initial value problem?

...ln|sec(t)| -2*ln|sin(t)| + C w = C*sec(t)/sin^2(t) = v' v = ∫ Csec (t)/sin^2(t)) dt = ∫Ccos(t)/(1 - sin^2(t))*(sin^2(t)) dt Let z...

1 Answers · Science & Mathematics · 30/09/2013

2. ### Solve the differential equation or initial-value problem?

...d/dx[ycos(x)] = cos(x) ycos(x) = sin(x) + C y = tan(x) + Csec (x) -------------- 2.) 1/√(x^2 - 3) (dy/dx) = xcos^2(y) sec^2(y) dy = x...

1 Answers · Science & Mathematics · 25/02/2013

3. ### Use implicit differentiation to find dy/dx?

dx+ csec ^2(xy).( xdy + ydx)=0 dy(x csec ^2(xy)=(-1-ycsec^2(xy))dx dy/dx=(-1-ycsec^2(xy))/(x csec ^2(xy)

4 Answers · Science & Mathematics · 28/10/2012

...2(x)]/cos(x)(x) - Dsin(x)[tan^2(x)] = Acos(x) - Bsin(x) + Csec (x) - Ccos(x) - Dsin(x)*[sec^2(x) - 1] = Acos(x) - Bsin...

1 Answers · Science & Mathematics · 05/05/2012

5. ### Mathematics: first-order linear ordinary differential equation?

...xsin(x) - ∫ sin(x)dx ycos(x) = xsin(x) + cos(x) + C y = xtan(x) + Csec (x) + 1 answer//

3 Answers · Science & Mathematics · 20/08/2011

6. ### Hard diff eq problem that I need help with, Please!?

y' - (tan x)y = 1 y ' cos x - (sin x) y = cos x y ' cos x +(cos x)' y = cos x (y cos x) ' = cos x y cos x = sin x +C y = (sin x +C)/cos x y = tan(x) + Csec (x)

1 Answers · Science & Mathematics · 09/03/2010

7. ### Is it possible to make a TI-84+ program where it wil appear in the math menu and not run as a program?

No. a TI-Basic program doesn't have any where near the capability of doing what you want. You would have to program it in Z80 Assembly, which is by far too tedious. If you have a cable that connects the TI-84 Plus to...

1 Answers · Science & Mathematics · 25/01/2010

8. ### Prove that tanx + cotx = (secx)(cscx).?

left part tan x = sinx / cosx cot x = cosx / sinx sin²x + cos²x = 1 ((sinx)² + (cosx)²) / (cosx*sinx) = 1 / (cosx* sinx) right part sec = 1 / cosx csec = 1 / sinx 1 / (sinx*cosx) so left = right

4 Answers · Science & Mathematics · 09/01/2010

...hence we can find perpendicular = 3 hence tan A = perpendicular/ base = -3/2 Csec is 1/ sinA and Sin A = p/h = 3/5sp Csec ...

4 Answers · Science & Mathematics · 30/08/2009

10. ### Can you help with this L'Hospital's Rule Problem?

Just keep taking the derivatives until it isn't 0/0, you're allowed to do that. With the x^3 on the bottom you'll need to apply L'Hopital's rule 3 times to get this not to be zero (x^3 -> 3x^2 -> 6x -> 6.)

1 Answers · Science & Mathematics · 22/04/2009