What is the solution for (d^2+9)y=sec(3x) ?

The solution for (d^2+9)y=sec(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/9 ln(cos(3x))cos(3x)+1/3 xsin(3x)

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(d^2+9)y=sec(3x)

(d^{2} + 9) y = sec (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{9} ln (cos (3 x)) cos (3 x) + \frac{1}{3} x sin (3 x)

Solution steps

(d^{2} + 9) y = sec (3 x)

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{9} ln (cos (3 x)) cos (3 x) + \frac{1}{3} x sin (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{1}{9} ln (cos (3 x)) cos (3 x) + \frac{1}{3} x sin (3 x)

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{t}

(d^{2} - 3 d + 2) \cdot y = e^{5 x}

y^{^{''}} - y = \frac{1}{x}

(d^{2} + 3 d + 2) y = x^{2} + sin (x)

y^{^{'''}} - y^{^{'}} = 4 e^{- x} + 3 e^{2 x}

What is the solution for (d^2+9)y=sec(3x) ?
The solution for (d^2+9)y=sec(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)+1/9 ln(cos(3x))cos(3x)+1/3 xsin(3x)