What is the solution for y^{''}+16y=t^2+1 ?

The solution for y^{''}+16y=t^2+1 is y=c_{1}cos(4t)+c_{2}sin(4t)+(t^2)/(16)+7/128

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y^{''}+16y=t^2+1

y^{^{''}} + 16 y = t^{2} + 1

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{t ^{2}}{16} + \frac{7}{128}

Solution steps

y^{''} (t) + 16 y = t^{2} + 1

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{t ^{2}}{16} + \frac{7}{128}

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{t ^{2}}{16} + \frac{7}{128}

y^{^{''}} - 2 y^{^{'}} + y = 3 x^{2} - 5

y^{^{''}} - 7 y^{^{'}} = t^{2}

y^{^{''}} - y^{^{'}} + 2 = 0

y^{^{''}} - a^{2} y = sech (a x)

y^{^{''}} - 5 y^{^{'}} + 6 y = - 4 e^{3 x} + 10 e^{x} - 12

What is the solution for y^{''}+16y=t^2+1 ?
The solution for y^{''}+16y=t^2+1 is y=c_{1}cos(4t)+c_{2}sin(4t)+(t^2)/(16)+7/128