Problem using variation of parameters

AI Thread Summary
The discussion focuses on solving the differential equation y'' + y = 2sec(t) + 3 using the method of variation of parameters. The user confirms the characteristic equation yields complex roots, leading to the fundamental solutions y1 = cos(t) and y2 = sin(t). They derive expressions for u1 and u2, which involve integrals of the non-homogeneous part, and seek verification of their calculations. The final solution is provided, confirming that the approach and calculations are correct. The overall process emphasizes the importance of integrating the specific functions derived from the differential equation to find the general solution.
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the problem is fin the general solution of the differential eq :

y''+y=2sect + 3 (-pi/2 < t < pi/2)

using variation of parameters.

I just needed a check to make sure my answer was correct.

r^2+1 = 0
r= -i
r= i
y1= cost
y2= sint
g(t)= 2sect+ 3

y(t) = c1cost + c2sint + Y(t)
Y(t) = u1y1 + u2y2
u1 = -(integal) (y2*g(t))/W in which W = 1
= -(int) sint(2sect+ 3)
= -(int) sint(2/cost+3)
= -(2 (int) tant + 3 (int) sint)
is this correct, where do i go from here

u2= (integral) y1*g(t)/ W
= (int) cost(2sect + 3)/ W
= (int) (2*(cost/cost) + 3cost)
= (int) 2 + (int)3 cost
= 2+3(sint)
is this correct

and then I plug these back into the Y(t) eq and add this to y(t)?
 
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Yup. And your answer is correct.
 
I don't know what u did,but here's where you have to get...

\frac{d^2y}{dx^2}+y=2\sec x+3

Exact solution is :

y\left( x\right) =2\ln \left( 1-\cos x-\sin x\right) \cos x+3\cos x+2\ln \left( 1-\cos x+\sin x\right) \cos x-4\ln \left( \sin x\right) \cos x
\allowbreak +2\ln \left( \cos x+1\right) \cos x-2\ln 2\cos x+4\sin x\arctan \left(\frac{\sin x}{\cos x+1}\right)+3+C_1\cos x+\allowbreak C_2\sin x

Daniel.
 
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