What is the solution for y^{''}-4y^'-21=0 ?

The solution for y^{''}-4y^'-21=0 is y=c_{1}e^{4t}+c_{2}-(21t)/4

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y^{''}-4y^'-21=0

y^{^{''}} - 4 y^{^{'}} - 21 = 0

y = c_{1} e^{4 t} + c_{2} - \frac{21 t}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 21 = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} - \frac{21 t}{4}

y = c_{1} e^{4 t} + c_{2} - \frac{21 t}{4}

y^{^{''}} + 256 y = sec (16 x)

y^{^{''}} + 7 y^{^{'}} + 6 y = sin (2 t), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} - 5 y = 22 + 13 x - 5 x^{2}

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

y^{^{''}} + y^{^{'}} = e^{t} \cdot cos (t)

What is the solution for y^{''}-4y^'-21=0 ?
The solution for y^{''}-4y^'-21=0 is y=c_{1}e^{4t}+c_{2}-(21t)/4