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

The solution for y^{'''}+y^'=sec^2(x) is y=c_{1}+c_{2}cos(x)+c_{3}sin(x)-cos(x)ln(sec(x)+tan(x))

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

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

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - cos (x) ln (sec (x) + tan (x))

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - cos (x) ln (sec (x) + tan (x))

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - cos (x) ln (sec (x) + tan (x))

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 9 + 2 sin (x)

y^{^{''}} - y = e^{x} sin (x)

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{(\frac{x}{2})}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - y = 20 x e^{x}

x^{^{''}} + 3 x^{^{'}} - 10 x = 6 e^{4 t}

What is the solution for y^{'''}+y^'=sec^2(x) ?
The solution for y^{'''}+y^'=sec^2(x) is y=c_{1}+c_{2}cos(x)+c_{3}sin(x)-cos(x)ln(sec(x)+tan(x))