What is the solution for y^{''}-4y^'+13(t)=0 ?

The solution for y^{''}-4y^'+13(t)=0 is y=c_{1}e^{4t}+c_{2}+(13t^2)/8+(13t)/(16)

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y^{''}-4y^'+13(t)=0

y^{^{''}} - 4 y^{^{'}} + 13 (t) = 0

y = c_{1} e^{4 t} + c_{2} + \frac{13 t ^{2}}{8} + \frac{13 t}{16}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 13 (t) = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} + \frac{13 t ^{2}}{8} + \frac{13 t}{16}

y = c_{1} e^{4 t} + c_{2} + \frac{13 t ^{2}}{8} + \frac{13 t}{16}

y^{^{''}} - y^{^{'}} = 2 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = 12 cos (3 t)

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

y^{^{'''}} + 3 y^{^{'}} = sin (x) + 1

y^{^{''}} - 4 y = 12 t^{2} cos^{2} (t)

What is the solution for y^{''}-4y^'+13(t)=0 ?
The solution for y^{''}-4y^'+13(t)=0 is y=c_{1}e^{4t}+c_{2}+(13t^2)/8+(13t)/(16)