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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int (\frac{1 + x + x}{x}) d x

\int \frac{x}{2 1 + x} d x

x = - y, x = 2 - y^{2}

x \to 6 lim (g (x))

\int \frac{1}{x ^{2} 7 - 16 x ^{2}} d x

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