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

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

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

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

Solution steps

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

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)

x \to 2 lim (x^{3} + 1)

\frac{d y}{d x} = \frac{x ^{2} - y ^{2} + y}{x}

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 3 y})

\int \frac{x ^{2} + x}{x ^{3} - x ^{2} + x - 1} d x

y^{^{'}} + \frac{1 - x}{2 x} y = 0

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