Problem using variation of parameters

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SUMMARY

The discussion centers on solving the differential equation y'' + y = 2sec(x) + 3 using the method of variation of parameters. The characteristic equation yields complex roots r = ±i, leading to the complementary solution y_c = c1*cos(t) + c2*sin(t). The particular solution Y(t) is derived using integrals involving the functions y1 = cos(t) and y2 = sin(t), with the final exact solution provided as a complex logarithmic expression. The calculations and methodology presented are confirmed to be correct.

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  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with the method of variation of parameters.
  • Knowledge of trigonometric functions and their integrals.
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Mathematics students, educators, and professionals working with differential equations, particularly those interested in advanced solution techniques and applications in physics and engineering.

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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...

[tex]\frac{d^2y}{dx^2}+y=2\sec x+3[/tex]

Exact solution is :

[tex]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[/tex]
[tex]\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[/tex]

Daniel.
 
Last edited:

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