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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = 4 sinh (x)

2 q^{^{''}} + 10 q^{^{'}} + 12 q = e^{- 3 t}

\frac{d ^{2} y}{d x ^{2}} + 12 \frac{d y}{d x} + 32 y = 160

y^{^{''}} - 4 y^{^{'}} = 16 t + 36, y (- 2) = 2, y^{^{'}} (- 2) = - 1

\frac{d ^{2} y}{d x ^{2}} + a \frac{d y}{d x} = b

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