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

The solution for y^{''}+y^'=tan^2(x) is y=c_{1}+c_{2}e^{-x}-x+tan(x)+e^{-x}*integral of-e^xtan^2(x)

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y^{''}+y^'=tan^2(x)

y^{^{''}} + y^{^{'}} = tan^{2} (x)

y = c_{1} + c_{2} e^{- x} - x + tan (x) + e^{- x} \cdot \int - e^{x} tan^{2} (x) d x

Solution steps

y^{''} (x) + y^{'} (x) = tan^{2} (x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - x + tan (x) + e^{- x} \cdot \int - e^{x} tan^{2} (x) d x

y = c_{1} + c_{2} e^{- x} - x + tan (x) + e^{- x} \cdot \int - e^{x} tan^{2} (x) d x

\frac{d ^{2} x}{d t ^{2}} + 81 x = 2 sin (9 t)

(D^{2} - 1) y = e^{- x} (2 sin (x) + 4 cos (x))

\frac{d ^{2} y}{d x ^{2}} - y = x \cdot e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 x + 4 - e^{5 x}

y^{^{''}} - 6 y^{^{'}} + 5 y = 34 sin (3 t)

What is the solution for y^{''}+y^'=tan^2(x) ?
The solution for y^{''}+y^'=tan^2(x) is y=c_{1}+c_{2}e^{-x}-x+tan(x)+e^{-x}*integral of-e^xtan^2(x)