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

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

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

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

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (x) + sec (x))

Solution steps

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

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (x) + sec (x))

y = c_{1} cos (x) + c_{2} sin (x) - cos (x) ln (tan (x) + sec (x))

t an g e n t f (x) = x^{2} + x, at x = 3

\frac{d}{d x} (- 7 x - 10)

\int 2^{2} d x

\frac{d}{d y} (e^{- 2 y})

\int \frac{ln ^{2} ( y )}{3 y + 8 y ln ^{3} ( y )} d y

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