What is the solution for y^{''}+y^'=tan(t) ?

The solution for y^{''}+y^'=tan(t) is y=c_{1}+c_{2}e^{-t}-ln(cos(t))+e^{-t}*integral of-e^ttan(t)

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y^{''}+y^'=tan(t)

y^{^{''}} + y^{^{'}} = tan (t)

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

Solution steps

y^{''} (t) + y^{'} (t) = tan (t)

Solve linear ODE:

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

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

x^{^{''}} - 2 x^{^{'}} + 2 x = t

y^{^{''}} - 4 y^{^{'}} - 2 y = 2 e^{- 3 t}, y (0) = 1, y^{^{'}} (0) = 7

y^{^{''}} + 4 y^{^{'}} + 12 y = 4 x^{2}

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 9 e^{3 x}

y^{^{''}} + 6 y^{^{'}} + 5 y = t \cdot e^{- t}

What is the solution for y^{''}+y^'=tan(t) ?
The solution for y^{''}+y^'=tan(t) is y=c_{1}+c_{2}e^{-t}-ln(cos(t))+e^{-t}*integral of-e^ttan(t)