What is the solution for x^{''}-6x^'+10x=30t ?

The solution for x^{''}-6x^'+10x=30t is x=e^{3t}(c_{1}cos(t)+c_{2}sin(t))+3t+9/5

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x^{''}-6x^'+10x=30t

x^{^{''}} - 6 x^{^{'}} + 10 x = 30 t

x = e^{3 t} (c_{1} cos (t) + c_{2} sin (t)) + 3 t + \frac{9}{5}

Solution steps

x^{''} (t) - 6 x^{'} (t) + 10 x = 30 t

Solve linear ODE:

x = e^{3 t} (c_{1} cos (t) + c_{2} sin (t)) + 3 t + \frac{9}{5}

x = e^{3 t} (c_{1} cos (t) + c_{2} sin (t)) + 3 t + \frac{9}{5}

y^{^{''}} - y^{^{'}} - 12 y = (x + 1) e^{x}

y^{^{''}} - 14 y^{^{'}} + 49 y = 35 x + 4

\frac{d ^{2} y}{d t ^{2}} + 4 y = sec^{2} (2 t)

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

y^{^{''}} + y = e^{t} + e^{- t}

What is the solution for x^{''}-6x^'+10x=30t ?
The solution for x^{''}-6x^'+10x=30t is x=e^{3t}(c_{1}cos(t)+c_{2}sin(t))+3t+9/5