Find the general solution of the DOE?

y''−2y'+5y=e˟tan2x … (i)

Simplify with sub y=ze˟ for which y’=e˟(z+z’) and y’’=e˟(z’’+2z’+z)

After substitution (i) reduces to z’’+4z = tan2x … (ii)

Now use Variation of Parameters applied to homogeneous solution of (ii)

So let z = Acos2x+Bsin2x where A and B are now functions of x

z’ = A’cos2x−2Asin2x+B’sin2x+2Bcos2x

Constrain A & B to satisfy A’cos2x+B’sin2x = 0 … (iii)

Then z’ = −2Asin2x+2Bcos2x and so z’’ = −2A’sin2x−4Acos2x + 2B’cos2x−4Bsin2x

With these subs (ii) reduces to −2A’sin2x+2B’cos2x = tan2x

Solve this with (iii) for A’ and B’ to get A’ = −½sin2xtan2x and B’ = ½cos2xtan2x

From these determine A & B

A’ = −½sin2xtan2x = −½(1−cos²2x)/cos2x = −½(sec2x−cos2x)

Integrate this for A = A₀ − ½(½logₑ(sec2x+tan2x) − ½sin2x) = A₀ − ¼logₑ(sec2x+tan2x) + ¼sin2x

B’ = ½cos2xtan2x = ½sin2x ⟹ B = B₀ − ¼cos2x

∴ general solution of (ii) is z = ( A₀−¼logₑ(sec2x+tan2x)+¼sin2x) cos2x + (B₀−¼cos2x) sin2x

This simplifies to z = A₀cos2x+B₀sin2x −¼cos2x.logₑ(sec2x+tan2x)

Hence General solutions of (i) is y = e˟{ A₀cos2x+B₀sin2x −¼cos2x.logₑ(sec2x+tan2x) }

y'' - 2y' + 5y = e^(x).tan(2x)

Characteristic equation

r² - 2r + 5 = 0

r² - 2r = - 5

r² - 2r + 1 = - 5 + 1

r² - 2r + 1 = - 4

r² - 2r + 1 = 4i²

(r - 1)² = (± 2i)²

r - 1 = ± 2i

r = 1 ± 2i

r₁ = 1 + 2i

r₂ = 1 - 2i

y = e^(x) * [A.cos(2x) + B.sin(2x)]

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