What is the solution for y^{''}+25y=25sec^2(5t) ?

The solution for y^{''}+25y=25sec^2(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)-1+ln(sec(5t)+tan(5t))sin(5t)

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y^{''}+25y=25sec^2(5t)

y^{^{''}} + 25 y = 25 sec^{2} (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) - 1 + ln (sec (5 t) + tan (5 t)) sin (5 t)

Solution steps

y^{''} (t) + 25 y = 25 sec^{2} (5 t)

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) - 1 + ln (sec (5 t) + tan (5 t)) sin (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) - 1 + ln (sec (5 t) + tan (5 t)) sin (5 t)

x^{''} + 2 x = 8, x (π) = 0, x (0) = 0

y^{^{''}} - y^{^{'}} - 42 y = - 122 t + 84 t^{2}

\frac{d ^{2} y}{d t ^{2}} + 12 \frac{d y}{d t} + 36 y = - 4 t + 3

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

y^{^{''}} + 2 y^{^{'}} = cos (t), y (0) = 0

What is the solution for y^{''}+25y=25sec^2(5t) ?
The solution for y^{''}+25y=25sec^2(5t) is y=c_{1}cos(5t)+c_{2}sin(5t)-1+ln(sec(5t)+tan(5t))sin(5t)