What is the general solution for solvefor x,(D^2-3D+2)y=sec^2(e^{-x}) ?

The general solution for solvefor x,(D^2-3D+2)y=sec^2(e^{-x}) is

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solvefor x,(D^2-3D+2)y=sec^2(e^{-x})

solve for

x, (D^{2} - 3 D + 2) y = sec^{2} (e^{- x})

x = - ln (arcsec (y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (- arcsec (y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (arcsec (- y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (- arcsec (- y (D^{2} - 3 D + 2)) + 2 πn)

Solution steps

(D^{2} - 3 D + 2) y = sec^{2} (e^{- x})

Switch sides

sec^{2} (e^{- x}) = (D^{2} - 3 D + 2) y

Solve by substitution

sec (e^{- x}) = (D^{2} - 3 D + 2) y, sec (e^{- x}) = - (D^{2} - 3 D + 2) y

sec (e^{- x}) = (D^{2} - 3 D + 2) y : x = - ln (arcsec (y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (- arcsec (y (D^{2} - 3 D + 2)) + 2 πn)

sec (e^{- x}) = - (D^{2} - 3 D + 2) y : x = - ln (arcsec (- y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (- arcsec (- y (D^{2} - 3 D + 2)) + 2 πn)

Combine all the solutions

x = - ln (arcsec (y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (- arcsec (y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (arcsec (- y (D^{2} - 3 D + 2)) + 2 πn), x = - ln (- arcsec (- y (D^{2} - 3 D + 2)) + 2 πn)

sin (θ) = \frac{7 sin ( 14 0 ^{\circ} )}{16.12288984}

- 5 cos^{2} (x) = 0

tan (2 x) + sec (2 x) = 5

tan (2 x) + sec (2 x) = 1

cos^{2} (x) = \frac{( 4 k - 5 )}{10}

What is the general solution for solvefor x,(D^2-3D+2)y=sec^2(e^{-x}) ?
The general solution for solvefor x,(D^2-3D+2)y=sec^2(e^{-x}) is