What is the solution for y^{''}-y=sec(x),y(0)=1,y^'(0)=0 ?

The solution for y^{''}-y=sec(x),y(0)=1,y^'(0)=0 is

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y^{''}-y=sec(x),y(0)=1,y^'(0)=0

y^{^{''}} - y = sec (x), y (0) = 1, y^{^{'}} (0) = 0

y = (1 - c_{2} - \int \frac{1}{2} d 0 - \int - \frac{1}{2} d 0) e^{x} + c_{2} e^{- x} + e^{x} \cdot \int \frac{e ^{- x} sec ( x )}{2} d x + e^{- x} \cdot \int - \frac{e ^{x} sec ( x )}{2} d x

Solution steps

y^{''} (x) - y = sec (x)

Solve linear ODE:

y = (1 - c_{2} - \int \frac{1}{2} d 0 - \int - \frac{1}{2} d 0) e^{x} + c_{2} e^{- x} + e^{x} \cdot \int \frac{e ^{- x} sec ( x )}{2} d x + e^{- x} \cdot \int - \frac{e ^{x} sec ( x )}{2} d x

y = (1 - c_{2} - \int \frac{1}{2} d 0 - \int - \frac{1}{2} d 0) e^{x} + c_{2} e^{- x} + e^{x} \cdot \int \frac{e ^{- x} sec ( x )}{2} d x + e^{- x} \cdot \int - \frac{e ^{x} sec ( x )}{2} d x

y^{^{''}} + 4 y = 4 sin^{2} (x)

y^{^{''}} - y = 2 e^{- t}

y^{^{''}} + 4 y = sin (x), y (0) = 1, y^{^{'}} (0) = 4

x^{^{''}} - x = 3 e^{- 2 t}

y^{^{''''}} - y = 7 t + cos (t)

What is the solution for y^{''}-y=sec(x),y(0)=1,y^'(0)=0 ?
The solution for y^{''}-y=sec(x),y(0)=1,y^'(0)=0 is