What is the solution for y^{''}+2y^'+y=e^{-x}sec^2(x) ?

The solution for y^{''}+2y^'+y=e^{-x}sec^2(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-e^{-x}ln(cos(x))

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} ln (cos (x))

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} ln (cos (x))

y = c_{1} e^{- x} + c_{2} x e^{- x} - e^{- x} ln (cos (x))

y^{^{''}} + 3 \cdot y^{^{'}} = 2

y^{^{''}} - y^{^{'}} = 0 + 2 t^{2} + t - 1

y^{^{''}} - y^{^{'}} + y = 25 sin (2 x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{( 1 - x ) ^{2}}

x^{^{''}} = - a^{2} x + b

What is the solution for y^{''}+2y^'+y=e^{-x}sec^2(x) ?
The solution for y^{''}+2y^'+y=e^{-x}sec^2(x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-e^{-x}ln(cos(x))